111662935-BINOMIAL-THEOREM-JEE-MAIN.pdf

Any algebraic expression which contains two dissimilar terms is called binomial expression. Any algebraic expression which contains two dissimilar terms is called binomial expression. For example : x + y, x For example : x + y, x22_{y +y +} 2 2 xy xy 1 1 , 3 , 3  – – x, x, _xx22

!!

₁₁ + + _{33 11 /  / 33} )) 1 1 x x (( 1 1

!!

 etc. etc. Factorial notation :

Factorial notation : _{or n! is pronounced as factorial n and is defined asor n! is pronounced as factorial n and is defined as}

n! = n! =

""

##

$$

%%

&

&

''

''

0 0 n n if if ;; 1 1 N N n n if if ;; 1 1 .. 2 2 .. 3 3 )... )... 2 2 n n )( )( 1 1 n n (( n n Note :

Note : n! = n . (nn! = n . (n – – 1)! 1)! ;; nn

&

& N

N

The term The term nn_CC

rrdenotes number of combinations of r things choosendenotes number of combinations of r things choosen

from n

from n distinct things mathematicallydistinct things mathematically,,nn_CC

rr== _{((nn rr)!)!rr!!}

!! n n

''

, n, n

&

&

N, rN, r

&

&

 W, 0 W, 0

))

rr

))

nn

Note :

Note : Other symbols of ofOther symbols of ofnn_CC

rr are are

 ₊

 ₊

****

 ,

 ,

----. 

. 

/ 

/ 

rr n n  and C(n, r).  and C(n, r). n n rr :: (i) (i) nn_CC rr = = n n_CC n n – – r r Note : Note : IfIf nn_CC x x== n n_CC y y

0

0

Either Either x x = = y y or or x x + + y y = = nn (ii) (ii) nn_CC rr + + n n_CC rr – – 1 1 = = n + 1 n + 1_CC rr (iii) (iii) 1 1 rr n n rr n n C C C C ''  = = rr 1 1 rr n n

''

!!

(iv) (iv) nn_CC rr == _rr n n _nn  –  –11_CC rr – –11 == rr((rr 11)) )) 1 1 n n (( n n

''

''

n n – –22_CC rr – –22 = = ... =... = rr((rr 11)()(rr 22)...)...22..11 )) )) 1 1 rr (( n n (( )... )... 2 2 n n )( )( 1 1 n n (( n n

''

''

''

''

''

''

((vv)) IIf n f n aand nd r r arare e rerelalatitivvelely y prpriimeme, t, thhenennn_CC

rr is divisible by n. But  is divisible by n. But converse is not converse is not necessarily true.necessarily true.

 (a + b)  (a + b)nn _= = nn_CC 0 0 a a n n_bb00 _+ + nn_CC 1 1 a a n n – –11bb11 + + nnCC 2 2 a a n n – –22bb22 +...+ +...+ nnCC rr a a n n – –rr b brr +... + +... + nnCC n n a a 0 0 _b bnn where n where n

&

&

 N N

o orr ((a a + + bb))nn _= =

1

1

%% '' n n 0 0 rr rr rr n n rr n n_{CC aa bb} Note :

Note : If we put a = 1 and b If we put a = 1 and b = x in the above binomial expansion, then= x in the above binomial expansion, then o orr ((1 1 + + xx))nn ₌₌ nn_CC 0 0 + + n n_CC 1 1 x  x ++ n n_CC 2 2 x x 2 2 _{+... + +... +} nn_CC rr x x rr _+...+ +...+ n n_CC n n x x n n o orr ((1 1 + + xx))nn ₌₌

1

1

%% n n 0 0 rr rr rr n n_{CC xx}

Binomial Theorem

Binomial Theorem

“

“ObviousObvious”” is the most dangerous word in mathematics... Bell, Eric Temple is the most dangerous word in mathematics... Bell, Eric Temple

Binomial expression : Binomial expression :

Terminology used in binomial theorem : Terminology used in binomial theorem :

Mathematical meaning of

Mathematical meaning of nn

rr

Properties related to Properties related to

Statement of binomial theorem : Statement of binomial theorem :

Regarding Pascal

Regarding Pascal’’s Triangle, we note the s Triangle, we note the following :following : (a

(a)) EaEach ch row row of of ththe e tritriananglgle e bebegigins ns wiwith th 1 a1 and nd enends ds wiwith th 1.1. (b

(b)) AnAny eny entry try in a rin a row iow is the ss the sum of tum of two ewo entrntrieies in ts in the phe precrecededing ing rowrow, one , one on thon the immee immediadiate lete left aft andnd the other on the imm

the other on the imm ediate right.ediate right. Example # 3 :

Example # 3 : The number of dissimilar terms in the expansion of (1The number of dissimilar terms in the expansion of (1 – – 3x + 3x 3x + 3x22 – – x x33))2020 is is ((AA))2211 ((BB)) 3311 ((CC))4411 ((DD))6611 Solution :

Solution : (1(1 – – 3x + 3x 3x + 3x22 – – x x33))2020 = [(1 = [(1 – – x) x)33]]2020 = (1 = (1 – – x) x)6060 Therefore number of dissimilar terms

Therefore number of dissimilar terms in the expansion of (1in the expansion of (1 – – 3x + 3x 3x + 3x22 – – x x33))2020 is is 61.61.

(x + y) (x + y)nn ₌₌ nn_CC 0 0 x x n n _y y00 _+ + nn_CC 1 1 xx n n – –11 _y y11 _{+ ...+ + ...+} nn_CC rr xx n n – –rr _y yrr _{+ ...+ + ...+} n n_CC n n x x 0 0 _y ynn (r + 1)

(r + 1)thth _{term is  term is called general term and denoted by Tcalled general term and denoted by T} r+1 r+1.. T T_r+1r+1 = = nn_CC rr x x n n – –rr _yyrr Note :

Note : The rThe rthth _{term from  term from the end is equal to the (nthe end is equal to the (n – – r + 2) r + 2)}thth _{term from the  term from the begining, begining, i.e.i.e.} nn_CC n

n – – r + 1 r + 1 xx

rr – – 1 1 _y ynn – – r + 1 r + 1

Example # 4 :

Example # 4 : FFiinndd ((ii)) 2288thth _{term of (5x + 8y) term of (5x + 8y)}3030 _{((iiii)) 77}thth _{term of term of}

9 9 2x 2x 5 5 5 5 4x 4x

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

_''

Solution : Solution : ((ii)) TT_{27 + 127 + 1} == 3030_CC 27 27 (5x) (5x) 30 30 – – 27 27 (8y) (8y)2727 == !! 27 27 !! 3 3 !! 30 30  (5x)  (5x)33 _{. (8y) . (8y)}2727

((iiii)) 77tth h tteerrm m ooff

9 9 x x 2 2 5 5 5 5 x x 4 4

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

_''

T T_{6 + 16 + 1} == 99_CC 6 6 6 6 9 9 5 5 x x 4 4 ''

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

66 x x 2 2 5 5

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

''

₌₌ !! 6 6 !! 3 3 !! 9 9 33 5 5 x x 4 4

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

66 x x 2 2 5 5

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

= = ₃₃ x x 10500 10500 Example # 5 :

Example # 5 : Find the number of rational terms in the expansion of (9Find the number of rational terms in the expansion of (91/41/4 _{+ 8 + 8}1/61/6₎₎10001000_..

Solution :

Solution : The general term in the expansion ofThe general term in the expansion of

2 2

11 /  / 44 11 /  / 66

33

10001000

8 8 9 9

!!

 is is T T_r+1r+1 == 10001000_CC rr rr 1000 1000 4 4 1 1 9 9 ''

**

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

rr 6 6 1 1 8 8

_**

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

 =  = 10001000_CC rr 22 rr 1000 1000 3 3 '' 2 2 rr 2 2 The above term will be rational if exponent of 3 and

The above term will be rational if exponent of 3 and 2 are integers2 are integers It means It means 2 2 rr 1000 1000

 '

 '

 and  and 2 2 rr  must be integers  must be integers The possible set of values of r

The possible set of values of r is {0, 2, 4, ..., 1000}is {0, 2, 4, ..., 1000} Hence, number of rational terms is 501

Hence, number of rational terms is 501

((aa)) If If n in is es eveven, n, ththerere ie is os onlnly oy one ne mimiddddle le tetermrm, w, whihich ch isis

th th 2 2 2 2 n n

**

 +

 +

 ,

 ,

--. 

. 

/ 

/  !!

 term.  term.

((bb)) If If n in is os odddd, t, thehere re arare te two wo mimiddddle le tetermrms, s, wwhihich ch araree

th th 2 2 1 1 n n

**

 +

 +

 ,

 ,

--. 

. 

/ 

/  !!

 and  and th th 1 1 2 2 1 1 n n

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

!!

!!

 terms.  terms. Example # 6

Example # 6 :: Find the middle term(s) in the expansion ofFind the middle term(s) in the expansion of

(i) (i) 14 14 2 2 2 2 x x 1 1

_**

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

''

_(ii)(ii) 9 9 3 3 6 6 a a a a 3 3

_**

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

''

General term : General term : Middle term s) : Middle term s) :

Case

-Case - WhenWhen

b b a a 1 1 1 1 n n

!!

!!

 is an integer (say m), then  is an integer (say m), then

((ii)) TT_r+1r+1 > T > T_rr wwhheenn r r < < mm ((r r = = 11, , 22, , 3 3 ..., , mm – – 1) 1) ii..ee.. TT₂₂ > T > T₁₁, T, T₃₃> T> T₂₂, ..., T, ..., T_mm > T > T_{mm – –11} ((iiii)) TT_r+1r+1 = T = T_rr wwhheenn r r = = mm ii..ee.. TT_m+1m+1 = T = T_mm ((iiiiii)) TT_r+1r+1 < T < T_rr wwhheenn r r > > mm ((r = r = m + m + 11, , m + m + 22, ., ... .. ...n )n ) ii..ee.. TT_m+2m+2 < T < T_m+1m+1 , T , T_m+3m+3 < T < T_m+2m+2 , ...T , ...T_n+1n+1 < T < T_nn Conclusion : Conclusion : When When b b a a 1 1 1 1 n n

!!

!!

 is an integer, say m, then T

 is an integer, say m, then TTT_mm and T and T_m+1m+1 will be numerically greatest terms (both terms are will be numerically greatest terms (both terms are equal in magnitude) equal in magnitude) Case Case -When When b b a a 1 1 1 1 n n

!!

!!

 is not an integer (Let its integral part be m), then  is not an integer (Let its integral part be m), then

((ii)) TT_r+1r+1 > T > T_rr wwhheenn r r << b b a a 1 1 1 1 n n

!!

!!

(r = 1, 2, 3,..., m (r = 1, 2, 3,..., m – –1, m)1, m) ii..ee.. TT₂₂ > T > T₁₁ , T , T₃₃ > T > T₂₂, ..., T, ..., T_m+1m+1 > T > T_mm ((iiii)) TT_r+1r+1 < T < T_rr when r >when r >

b b a a 1 1 1 1 n n

!!

!!

(r = m + 1, m + 2, ...n) (r = m + 1, m + 2, ...n) ii..ee.. TT_m+2m+2 <  < TT_m+1m+1 ,  , TT_m+3m+3 < T < T_m+2m+2 , ..., T , ..., T_{n +1n +1} <  < TT_nn Conclusion : Conclusion : When When b b a a 1 1 1 1 n n

!!

!!

 is not an integer and its integral part is m, then T

 is not an integer and its integral part is m, then T_m+1m+1 will be the  will be the numerically greatestnumerically greatest

term. term. Note :

Note : (i(i)) In aIn any bny bininomiomial eal expxpanansisionon, th, the mide middldle tee term(rm(s) has) has grs greaeatetest bist binonomiamial col coefefficficieientnt.. In the expansion of (a + b)

In the expansion of (a + b)nn

IIff nn NNoo. . oof f ggrreeaatteesst t bbiinnoommiiaal l ccooeeffffiicciieennt t GGrreeaatteesst t bbiinnoommiiaal l ccooeeffffiicciieenntt E Evveenn 11 nn_CC n/2 n/2 O Odddd 22 nn_CC (n (n – – 1)/2 1)/2 andand n n_CC (n + 1)/2 (n + 1)/2 (V

(Values of both these alues of both these coefficients are equal )coefficients are equal ) (i

(ii)i) In oIn orderder to or to obtabtain tin the the term herm haviaving nung numerimericalcally ly greagreatestest cot coeffiefficiecient, nt, put put a = b = 1a = b = 1, an, and prod proceeceedd as discussed above.

as discussed above. Example # 8 :

Example # 8 : Find the numerically greatest term in the expansion of (3Find the numerically greatest term in the expansion of (3 – – 5x) 5x)1515 when x = when x =

5 5 1 1 .. Solution :

Solution : Let rLet rthth _{and (r + 1) and (r + 1)}thth _{be two consecutive terms in the expansion of ( be two consecutive terms in the expansion of ( 33 – – 5x) 5x)}1515

T T_{r + 1r + 1}

44

 T T_rr 15 15_CC rr 3 3 15 15 – – r r _{(| (| – – 5x|) 5x|)}rr

44

1515_CC rr – – 1 1 3 3 15 15 – – (r (r – – 1) 1) _{(| (| – – 5x|) 5x|)}rr – – 1 1 !! rr !! )) rr 15 15 (( )! )! 15 15

''

22

 |  | – – 5x | 5x |

44

!! )) 1 1 rr (( !! )) rr 16 16 (( )! )! 15 15 .. 3 3

''

''

22

5 . 5 . 5 5 1 1  (16  (16 – – r) r)

44

3r3r 16 16 – – r r

44

 3r 3r 4r 4r

))

 1 166 rr

))

 4 4

Self practice problems : Self practice problems :

((88)) If If n in is a s a poposisititive ve inintetegegerr, t, thehen sn shohow w ththat at 332n + 12n + 1 _{+ 2 + 2}n + 2n + 2 _{is divisible by 7. is divisible by 7.}

((99)) W hW haat t iis s tthhe e rreem am aiinnddeer wr whheen n 77103103 _{is divided by 25 . is divided by 25 .}

(1

(10)0) FinFind the d the laslast digt digit, lit, last tast two dwo digiigits ants and lad last thst three dree digiigits of thts of the nue number (mber (81)81)25.25.

((1111)) WhWhiich ch nunumbmber er iis ls larargger er (1(1..22))40004000 _{or 800 or 800}

Answers :

Answers : ((99)) 1188 ((1100)) 11, , 0011, , 000011 ((1111)) ((11..22))40004000_..

(i)

(i) Consider the expansionConsider the expansion

(x + y) (x + y)nn _= =

1

1

%% n n 0 0 rr rr n n_C C  x xnn – –rr _y yrr ₌₌ nn_CC 0 0 x x n n _y y00 _+ + nn_CC 1 1 xx n n – –11 _y y11 _{+ ...+ + ...+} nn_CC rr xx n n – –rr _yyrr _{+ ...+ + ...+} n n_CC n n x x 0

0 _y ynn  _{....(i)  ....(i)}

(ii)

(ii) Now replace yNow replace y

5

5

 – – y we get y we get

(x (x – – y) y)nn = =

1

1

%% n n 0 0 rr rr n n_C C (( – – 1) 1) r r x xnn – –rr _y yrr ₌₌ nn_CC 0 0 x x n n _y y00 _– – nn_CC 1 1 x x n n – –11 _y y11 _{+ ...+ + ...+} nn_CC rr ( ( – –1)1) rr_xxnn – –rr _y yrr _{+ ...+ + ...+} n n_CC n n ( ( – – 1) 1) n

n _x x00 _y ynn _{....(ii)....(ii)}

(iii)

(iii) Adding (i) & (ii), we getAdding (i) & (ii), we get (x + y) (x + y)nn _{+ (x + (x – – y) y)}nn _{=  = 2[2[}nn_CC 0 0xx n n _y y00₊₊ nn_CC 2 2 x x n n – – 2 2 _yy22_+...]+...] (iv)

(iv) Subtracting (ii) from (i), we getSubtracting (ii) from (i), we get (x + y) (x + y)nn _{– – (x (x  – – y) y)}nn _{=  = 2[2[}nn_CC 1 1xx n n – – 1 1 _y y11₊₊ nn_CC 3 3 x x n n – – 3 3 _yy33_+...]+...] (1 + x) (1 + x)nn _{=  = CC} 0 0 +  + CC11x + x + CC22xx 2 2 _{+ ... + C + ... + C} rrxx rr _{+ ... + C + ... + C} n nxx n n _...(1)...(1) where C

where C_rr denotes denotes nn_CC rr

(1

(1)) The The susum of m of ththe be bininomiomial al cocoefeffificicienents ts in in ththe ee expxpanansision on of of (1 + (1 + x)x)nn _{is 2 is 2}nn

Putting x = 1 in (1) Putting x = 1 in (1)  n  n_CC 0 0 + + n n_CC 1 1 + + n n_CC 2 2 + ...+ + ...+ n n_CC n n = 2 = 2 n n _...(2)...(2) or or

1

1

%%

%%

n n 0 0 rr n n rr n n_{CC 22}

((22)) AAggaaiin n ppuuttttiinng g x x == – –1 in (1), we get1 in (1), we get

 n  n_CC 0 0 – – n n_CC 1 1 + + n n_CC 2 2 – – n n_CC 3 3 + ... + ( + ... + ( – –1)1) n n nn_CC n n == 00 ...((33)) or or

1

1

%%

%%

''

n n 0 0 rr rr n n rr _{CC 00} )) 1 1 (( (3)

(3) The The sum osum of the f the binbinomiaomial col coeffiefficiecients nts at oat odd pdd posiositiotion is n is equequal tal to tho the sue sum of thm of the bie binominomial cal coeffioefficiecientsnts at even position and each is equal to 2

at even position and each is equal to 2nn – –11_..

from (2) and (3) from (2) and (3)  n  n_CC 0 0 + + n n_CC 2 2 + + n n_CC 4 4 + ... = + ... = n n_CC 1 1 + + n n_CC 3 3 + + n n_CC 5 5 + ... = 2 + ... = 2 n n – –11

((44)) SuSum om of tf two wo coconsnsececuutitive ve bibinonomimial al cocoefeffificicienentsts

n n_CC rr + + n n_CC rr – –11 == n+1 n+1_CC rr

0

0

LL..HH..SS. . == n n_CC rr + + n n_CC rr – –11 == _{((nn rr)!)!rr!!} !! n n

''

 + + ((nn rr 11)!)!((rr 11)!)! !! n n

''

!!

''

= = _((nn

_''

_rr)!)!nn!!_((rr

_''

_11)!)!

_99::

;;

11_rr

!!

_nn

_''

11_rr

_!!

₁₁

88

₆₆₇₇

 = = _((nn

_''

_rr)!)!nn!!_((rr

_''

_11)!)! )) 1 1 rr n n (( rr )) 1 1 n n ((

!!

''

!!

 =  = _((nn((

_''

nn_rr

!!

_!!

11₁₁)!)!_)!)!rr!! = = n+1n+1_CC rr = = R.H.S.R.H.S.

Some standard expansions : Some standard expansions :

Properties of binomial coefficients : Properties of binomial coefficients :

Method : By Integration Method : By Integration (1 + x) (1 + x)nn _{=  = CC} 0 0 +  + CC11x + x + CC22xx 2 2 _{+ ... + C + ... + C} n n x x n n_..

Integrating both sides, within the limits

Integrating both sides, within the limits – – 1 to 0. 1 to 0.

0 0 1 1 1 1 n n 1 1 n n )) x x 1 1 (( '' !!

66

66

77

88

99

99

::

;;

!!

!!

 =  = 0 0 1 1 1 1 n n n n 3 3 2 2 2 2 1 1 0 0 1 1 n n x x C C ... ... 3 3 x x C C 2 2 x x C C x x C C '' !!

66

66

77

88

99

99

::

;;

!!

!!

!!

!!

!!

1 1 n n 1 1

!!

 – – 0 = 0 0 = 0 – –

66

77

88

99

::

;;

!!

''

!!

!!

''

!!

''

!! 1 1 n n C C )) 1 1 (( ... ... 3 3 C C 2 2 C C C C₀₀ 11 22 nn 11 nn C C₀₀ – – 2 2 C C₁₁  +  + 3 3 C C₂₂  –  – ... + ( ... + ( – – 1) 1)nn 1 1 n n C C_nn

!!

 = = nn 11 1 1

!!

ProvedProved Example # 14 : Example # 14 : If (1 + x)If (1 + x)nn _{= C = C} 0 0 + C + C11x + x + CC22xx 2 2 _{+ ...+ C + ...+ C} n nxx n

n_{, then prove that, then prove that}

((ii)) CC₀₀22 _{+ C + C} 1 1 2 2 _{+ C + C} 2 2 2 2 _{+ ... + C + ... + C} n n 2 2 _= = 2n2n_CC n n ((iiii)) CC₀₀CC₂₂ + C + C₁₁CC₃₃ + C + C₂₂CC₄₄ + ... + C + ... + C_{nn – – 2 2} C C_nn = = 2n2n_CC n n – – 2 2oror 2n 2n_CC n + 2 n + 2 ((iiiiii)) 11. . CC₀₀22 _{+ 3 . C + 3 . C} 1 1 2 2 _{+ 5. C + 5. C} 2 2 2 2 _{+ ... + (2n + 1) . C + ... + (2n + 1) . C} n n 2 2 _{. = 2n. . = 2n.} 2n2n – – 1 1_CC n n + + 2n 2n_CC n n.. Solution : Solution : ((ii)) ((1 1 + + xx))nn _{=  = CC} 0 0 +  + CC11x + x + CC22xx 2 2 _{+ ... + C + ... + C} n n x x n n_{.. ...((ii))} (x + 1) (x + 1)nn _{= C = C} 0 0xx n n _{+ C + C} 1 1xx n n – – 1 1_{+ C+ C} 2 2xx n n – – 2 2 _{+ ... + C + ... + C} n nxx 0 0 _{...(ii)...(ii)} Multiply

Multiplying (i) aning (i) an d (ii)d (ii) (C (C₀₀ + C + C₁₁x + x + CC₂₂xx22 _{+ ... + C + ... + C} n nxx n n_{) () (CC} 0 0xx n n _{+ C + C} 1 1xx n n – – 1 1 _{+ ... + C + ... + C} n nxx 0 0_{) = (1 + x)) = (1 + x)}2n2n Comparing coefficient of x Comparing coefficient of xnn_,, C C₀₀22 _{+ C + C} 1 1 2 2 _{+ C + C} 2 2 2 2 _{+ ... + C + ... + C} n n 2 2 _= =2n2n_CC n n (i

(ii)i) FrFrom om ththe pe proroduduct ct of of (i(i) a) and nd (i(ii) i) cocompampariring ng cocoefeffificicienents ts of of xxnn – – 2 2 _{or x or x}n + 2n + 2 _{both sides, both sides,}

C C₀₀CC₂₂ + C + C₁₁CC₃₃ + C + C₂₂CC₄₄ + ... + C + ... + C_{nn – – 2 2}CC_nn = = 2n2n_CC n n – – 2 2oror 2n 2n_CC n + 2 n + 2.. (iii)

(iii) Method : By SummationMethod : By Summation L.H.S. L.H.S. = = 1. 1. CC₀₀22 _{+ 3. C + 3. C} 1 1 2 2 _{+ 5. C + 5. C} 2 2 2 2 _{+ ... + (2n + 1)  + ... + (2n + 1) CC} n n 2 2_.. = =

1

1

%%

!!

n n 0 0 rr )) 1 1 rr 2 2 (( nn_CC rr 2 2 ₌₌

1

1

%% n n 0 0 rr rr .. 2 2 . (. (nn_CC rr)) 2 2 _+ +

1

1

%% n n 0 0 rr 2 2 rr n n_{CC ))} (( = 2 = 2

1

1

%% n n 1 1 rr n n .. .. nn – – 1 1_CC rr – – 1 1 n n_CC rr + + 2n 2n_CC n n (1 + x) (1 + x)nn _= = nn_CC 0 0 + + n n_CC 1 1 x  x ++ n n_CC 2 2 x x 2 2 _{+ ... + ...}nn_CC n n x x n n _...(i)...(i) (x + 1) (x + 1)nn – – 1 1 = = nn – – 1 1CC 0 0 x x n n – – 1 1 + + nn – – 1 1CC 1 1 x x n n – – 2 2 + ...+ + ...+nn – – 1 1CC n n – – 1 1xx 0 0 _{...(ii)...(ii)} Multiply

Multiplying (i) and (ii) ing (i) and (ii) and comparing coeffcients of and comparing coeffcients of xxnn_.. n n – – 1 1_CC 0 0 . . n n_CC 1 1 + + n n – – 1 1_CC 1 1 . . n n_CC 2 2 + ... + + ... + n n – – 1 1_CC n n – – 1 1.. n n_CC n n == 2n 2n – – 1 1_CC n n

1

1

%% '' '' n n 0 0 rr 1 1 rr 1 1 n n C C ..nn_CC rr = = 2n 2n – – 1 1_CC n n Hence,

Hence, required required summation summation is is 2n.2n.2n2n – – 1 1_CC n n ++ 2n 2n_CC n n = R.H.S. = R.H.S. Method : By Differentiation Method : By Differentiation (1 + x (1 + x22₎₎nn _{=  = CC} 0 0 +  + CC11xx 2 2 _{+  + CC} 2 2xx 4 4 _{+ C + C} 3 3xx 6 6 _{+ ...+ C + ...+ C} n n x x 2n 2n

Multiplying both sides by x Multiplying both sides by x x(1 + x x(1 + x22₎₎nn _{=  = CC} 0 0x + x + CC11xx 3 3 _{+  + CC} 2 2xx 5 5 _{+ ... + C + ... + C} n nxx 2n + 1 2n + 1_..

Differentiating both sides Differentiating both sides x . n (1 + x x . n (1 + x22₎₎nn – – 1 1 _{. 2x + (1 + x . 2x + (1 + x}22₎₎nn _{= C = C} 0 0 + 3. C + 3. C11xx 2 2 _{+ 5. C + 5. C} 2 2 x x 4 4 _{+ ...+ (2n + 1) C + ...+ (2n + 1) C} n n x x 2n 2n_...(i)...(i) (x (x22 _{+ 1 + 1))}nn _{=  = CC} 0 0 x x 2n 2n _{+  + CC} 1 1 x x 2n 2n – – 2 2 + C + C 2 2 x x 2n 2n – – 4 4 + ... + C + ... + C n n ...(ii)...(ii)

Multiplying (i) & (ii) Multiplying (i) & (ii) (C (C₀₀ + 3 + 3CC₁₁xx22 _{+ 5 + 5CC} 2 2xx 4 4 _{+ ... + (2n + 1) C + ... + (2n + 1) C} n n x x 2n 2n_{) () (CC} 0 0 x x 2n 2n _{+ C + C} 1 1xx 2n 2n – – 2 2 _{+ ... + C + ... + C} n n)) = 2n x = 2n x22 _{(1 + x (1 + x}22₎₎2n2n – – 1 1 + (1 + x + (1 + x22))2n2n comparing coefficient of x comparing coefficient of x2n2n_,, C C₀₀22 _{+ 3 + 3CC} 1 1 2 2 _{+ 5C + 5C} 2 2 2 2 _{+ ...+ (2n + 1) C + ...+ (2n + 1) C} n n 2 2_{= 2n .= 2n .} 2n2n – – 1 1_CC n n – – 1 1 + + 2n 2n_CC n n.. C C₀₀22 _{+ 3 + 3CC} 1 1 2 2 _{+ 5C + 5C} 2 2 2 2 _{+ ...+ (2n + 1) C + ...+ (2n + 1) C} n n 2 2_{= 2n .= 2n .}2n2n – –11_CC n n + + 2n 2n_CC n n. Proved. Proved

As we know the Binomial Theorem As we know the Binomial Theorem – – (x + y) (x + y)nn _= =

1

1

%% n n 0 0 rr rr n n_CC x xnn – –rr _yyrr ₌₌

1

1

%%

''

n n 0 0 rr ((nn rr)!)!rr!! !! n n  x  xnn – –rr _yyrr putting n putting n – – r  r = r= r₁₁ , r = r , r = r₂₂

tthheerreeffoorree,, ((x x + + yy))nn _= =

1

1

%% !!rr nn rr11 22 rr11!! rr22!! !! n n 2 2 1 1 rr rr _..yy x x T

Total number of termotal number of term s in the expansion of (s in the expansion of ( x + y)x + y)nn _{is equal to number  is equal to number of non-negative integral solutionof non-negative integral solution}

of r of r₁₁ + r + r₂₂ =  = nn ii..ee.. n+2n+2 – –11_CC 2 2 – –11 == n+1 n+1_CC 1 1 = = n + n + 11

In the same fashion we can write the multinomial theorem In the same fashion we can write the multinomial theorem (x (x₁₁ +  + xx₂₂ +  + xx₃₃ + ... x + ... x_kk))nn _= =

1

1

%% !! !! !!rr ... rr nn rr11 22 kk rr11!!rr22!...!...rrkk!! !! n n _{11 22 rrkk} k k rr 2 2 rr 1 1 ..xx ...xx x x Here total number of terms in the expansion of (x

Here total number of terms in the expansion of (x₁₁ +  + xx₂₂ + ... + x + ... + x_kk))nn _{is equal to number of non- is equal to number of}

non-negative integral solution of r

negative integral solution of r₁₁ + r + r₂₂ + ... + r + ... + r_kk = n = n ii..ee.. n+kn+k – –11_CC k k – –11

Example # 17 :

Example # 17 : Find the coefficient of aFind the coefficient of a22_bb33_cc44_{d in the expansion of (ad in the expansion of (a – – b b – – c + d) c + d)}1010

Solution : Solution : (a(a – – b b – – c + d) c + d)1010 ==

1

1

%% !! !! !!rr rr rr 1010 rr11 22 33 44 rr11!!rr22!!rr33!!rr44!! )! )! 10 10 (( 4 4 3 3 2 2 1 1 rr rr rr rr _{(( bb)) (( cc)) ((dd))} )) a a ((

''

''

we want to get a

we want to get a22 _b b33 _c c44 _{d  d tthhiis s iimmpplliiees s tthhaatt rr} 1 1 = 2, r = 2, r22 = 3, r = 3, r33 = 4, r = 4, r44 = 1 = 1

<

<

coeff. of acoeff. of a22 _b b33 _c c44 _{d is d is} !! 1 1 !! 4 4 !! 3 3 !! 2 2 )! )! 10 10 ((  (  ( – –1)1)33 ( ( – –1)1)44 == – – 12600 12600 Example # 18

Example # 18 :: In the expansion of In the expansion of

11 11 x x 7 7 x x 1 1

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

!!

!!

, find the term independent of x., find the term independent of x.

Solution : Solution : 11 11 x x 7 7 x x 1 1

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

!!

!!

 _= =

1

1

%% !! !!rr rr 1111 rr11 22 33 rr11!!rr22!!rr33!! )! )! 11 11 (( 33 2 2 1 1 rr rr rr x x 7 7 )) x x (( )) 1 1 ((

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

The exponent 11 is to be divided among the b

The exponent 11 is to be divided among the b ase variables 1, x andase variables 1, x and x x 7 7

 in such a way so that we  in such a way so that we get x

get x00_..

Therefore, possible set of values of (r

Therefore, possible set of values of (r₁₁, r, r₂₂, r, r₃₃) are (11, 0, 0), (9, 1, 1), (7, 2, 2), () are (11, 0, 0), (9, 1, 1), (7, 2, 2), ( 5, 3, 3), (3, 4, 4),5, 3, 3), (3, 4, 4), (1, 5, 5)

(1, 5, 5)

Hence the required term is Hence the required term is

)! )! 11 11 (( )! )! 11 11 ((  (7  (700_{) +) +} !! 1 1 !! 1 1 !! 9 9 )! )! 11 11 ((  7  711 _+ + !! 2 2 !! 2 2 !! 7 7 )! )! 11 11 ((  7  722 _+ + !! 3 3 !! 3 3 !! 5 5 )! )! 11 11 ((  7  733 ₊₊ !! 4 4 !! 4 4 !! 3 3 )! )! 11 11 ((  7  744 _+ + !! 5 5 !! 5 5 !! 1 1 )! )! 11 11 ((  7  755 = 1 + = 1 + ₉₉((1111_!!22)!)!_!!  . . ₁₁22_!!11!!_!!  7 711 _+ + !! 4 4 !! 7 7 )! )! 11 11 ((  .  . ₂₂44_!!22!! _!!  7 722 _+ + !! 6 6 !! 5 5 !! )) 11 11 ((  .  . ₃₃66_!!33!! _!!  7 733 + + ₃₃((1111_!!88))!!_!!  . . ₄₄88_!!44!! _!! 7 744 _+ + !! 10 10 !! 1 1 !! )) 11 11 ((  .  . ₅₅((1010_!!55))!!_!!  7 755 = 1 + = 1 +1111_CC 2 2 . . 2 2_CC 1 1 . 7 . 7 1 1 _+ +1111_CC 4 4 . . 4 4_CC 2 2 . 7 . 7 2 2 _+ +1111_CC 6 6 . . 6 6_CC 3 3 . 7 . 7 3 3 _+ +1111_CC 8 8 . . 8 8_CC 4 4 . 7 . 7 4 4 ₊₊ 1111_CC 10 10 . . 10 10_CC 5 5 . 7 . 7 5 5 = 1 + = 1 +

1

1

%% 5 5 1 1 rr rr 2 2 11 11_CC  .  . 2r2r_CC rr .  . 77 rr Multinomial theorem : Multinomial theorem :

Example-20 :

Example-20 : If x is so small such that its square and higher powers maIf x is so small such that its square and higher powers ma y be neglected, then find the value ofy be neglected, then find the value of

2 2  /   /  1 1 3 3  /   /  5 5 2 2  /   /  1 1 )) x x 4 4 (( )) x x 1 1 (( )) x x 3 3 1 1 ((

!!

''

!!

''

Solution : Solution : 11 /  / 22 3 3  /   /  5 5 2 2  /   /  1 1 )) x x 4 4 (( )) x x 1 1 (( )) x x 3 3 1 1 ((

!!

''

!!

''

 =  = 11 /  / 22 4 4 x x 1 1 2 2 3 3 x x 5 5 1 1 x x 2 2 3 3 1 1

**

 +

 +

 ,

 ,

--. 

. 

/ 

/  !!

''

!!

''

 =  = 2 2 1 1

_**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

_''

x x 6 6 19 19 2 2 2 2  /   /  1 1 4 4 x x 1 1 ''

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

!!

= = 2 2 1 1

_**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

''

xx 6 6 19 19 2 2

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

''

8 8 x x 1 1  _= = 2 2 1 1

_**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

''

''

xx 6 6 19 19 4 4 x x 2 2  _{=  = 11 – –} 8 8 x x  –  – 12 12 19 19 x x = = 11 – – 24 24 41 41  x  x Self practice problems :

Self practice problems : (1

(16)6) FinFind thd the poe possissiblble see set of vt of valualues oes of x fof x for wr whihich ech expaxpansinsion oon of (3f (3 – – 2x)2x)1/21/2 is valid in ascending is valid in ascending powers of x. powers of x. ((1177)) IIf f y y == 5 5 2 2  +  + 11₂₂..33_!! 2 2 5 5 2 2

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

 +  + !! 3 3 5 5 .. 3 3 .. 1 1 33 5 5 2 2

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

 + ..., then find the value of

 + ..., then find the value of yy22 _{+  + 2y2y}

((1188)) TThhe ce cooeeffffiicciieennt ot of xf x100100 _in in

2 2 )) x x 1 1 (( x x 5 5 3 3

''

''

 is  is ((AA) ) 110000 ((BB)) – –5757 (C)(C) – –197197 (D) 53(D) 53 Answers :

Answers : ((1166)) xx

&

&

**

 +

 +

 ,

 ,

--. 

. 

/ 

/ 

_''

2 2 3 3 ,, 2 2 3 3 ((1177)) 44 ((1188)) CC

20.

20. Find the coefficient of the term independent of x in the expansion ofFind the coefficient of the term independent of x in the expansion of

10 10 2 2  /   /  1 1 3 3  /   /  1 1 3 3  /   /  2 2 _{xx xx} 1 1 x x 1 1 x x x x 1 1 x x

_!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

''

''

''

((

''

((

21.

21. If in the expansion of (1 + x)If in the expansion of (1 + x)mm_(1(1 –

 –x)x)nn, the coefficients of x and x, the coefficients of x and x22 are 3 and – are 3 and –6 respectively. 6 respectively. Then find Then find the valuethe value of m.

of m. 22.

22. Find the number of terms in the expansion of (1 Find the number of terms in the expansion of (1 + 5+ 5 ₂₂ x)x)99 _{+ (1 + (1 –}

 – 5 5 22x)x)99.. 23.

23. If the coefficients of second, third and fourth If the coefficients of second, third and fourth terms in the expansion of (1 terms in the expansion of (1 + x)+ x)nn _{are in A.P., then  are in A.P., then find the find the valuevalue}

of n. of n. 24.

24. If in the expansion of (1If in the expansion of (1 – – x) x)2n2n – –11 the coefficient of x the coefficient of xrr is denoted by a is denoted by a_rr, then prove that a, then prove that a_{rr – –1}1 + a + a2n2n – –r r  =  = 00

25.

25. Using binomial theorem, prove that 2Using binomial theorem, prove that 23n3n

 –

 – 7n 7n – – 1 is divisible by 49 where  1 is divisible by 49 where nn

)

)

 N. N. 26.

26. Using binomial theorem, prove that 3Using binomial theorem, prove that 32n+22n+2

 –

 – 8n 8n – – 9 is divisible by 64, n 9 is divisible by 64, n

)

)

 N. N. 27.

27. Prove thatProve that

5 5 .. 4 4 1 1  –  – 4 4 .. 3 3 1 1 3 3 .. 2 2 1 1  –  – 2 2 .. 1 1 1 1

₍₍

 + ...  + ...

*

*

= = loglog_ee

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

e e 4 4 .. 28.

28. Find the sum of the infinite series 1 +Find the sum of the infinite series 1 +

!! 6 6 1 1 !! 4 4 1 1 !! 2 2 1 1

₍₍

₍₍

 + ...  + ... 29.

29. Prove that (xProve that (x22 – – y y22) +) + _22!! 1 1 (x (x44 – – y y44) +) + _33!! 1 1 (x (x66 – – y y66) + ... to) + ... to

*

*

 = = 2 2 2 2 y y x x _ee  –  – e e T

Tyyppe e ((IIVV) ) : : VVeerry y LoLonng g AAnnsswweer r TTyyppe e QQuueessttiioonnss:: [[006 6 MMaarrk k EEaacchh]] 30.

30. Find the value ofFind the value of

+

+

,,

((

((

''

n n 0 0 rr rr n n e e e e rr n n rr )) 10 10 log log 1 1 (( 10 10 log log rr 1 1 C C )) 1 1 (( .. 31.

31. If the coefficient of rIf the coefficient of rthth_{, (r + 1), (r + 1)}thth_{and (r + 2)and (r + 2)}thth  _{terms in the expansion of (1 + x)  terms in the expansion of (1 + x)}1414  _{are in A.P, then find  are in A.P, then find}

the value of r. the value of r. 32.

32. If the coefficients of three cosecutive termIf the coefficients of three cosecutive terms in the expansion of (1 + x)s in the expansion of (1 + x)nn _{are in the ratio 1 : 7 : 42. Find n are in the ratio 1 : 7 : 42. Find n}

33.

33. If 3If 3rdrd_{, 4, 4}thth_{, 5, 5}thth _{and 6 and 6}thth _{terms in the expansion of (x + terms in the expansion of (x +}

-

-

₎₎nn _{be respectively a, b, c and d  be respectively a, b, c and d then prove thatthen prove that}

bd bd c c ac ac b b 2 2 2 2

''

''

 =  = c c 3 3 a a 5 5 34.

34. If coefficients of three consecutive terms If coefficients of three consecutive terms in the expansion of (1 + x)in the expansion of (1 + x)nn _{be 76,95 and 76. Then find n. be 76,95 and 76. Then find n.}

35.

35. If the 2nd, 3rd and 4th terms in the expansion of (x + a)If the 2nd, 3rd and 4th terms in the expansion of (x + a)nn _{are 240,  are 240, 720 and 1080 respectively720 and 1080 respectively, find x, , find x, a and n.a and n.}

36.

36. Sum the series from n = 1 to n =Sum the series from n = 1 to n =

*

*

 , whose nth term is , whose nth term is (i)

(i) _((nn

₍₍

11_11))!! (ii)(ii) _((nn

₍₍

11_22))!! (iii)(iii) _11))!!  –  – n n 2 2 (( 1 1 37.

37. Prove thatProve that

log

log_ee

$$

& 

& 

_% 

_% 

mm_nn

 #

 #

 _"

 _"

!!

 = 2 = 2

_..

..

//

00

11

11

22

33

((

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

((

((

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

((

((

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

((

mm nn ... n n  –  – m m 5 5 1 1 n n m m n n  –  – m m 3 3 1 1 n n m m n n  –  – m m 33 55 38.

38. Prove thatProve that log

log_ee

$$

&  ((

_% 

_% 

& 

xx_xx 11

 #

 #

 _"

 _"

!!

 = =

..

//

00

11

22

33

((

((

((

((

((

((

_{55((22xx 11))} ... 1 1 )) 1 1 x x 2 2 (( 3 3 1 1 )) 1 1 x x 2 2 (( 1 1 2 2 _{33 55}

A-12.

A-12. The co-efficient of x in the expansion of (1The co-efficient of x in the expansion of (1

''

 2 2 xx33 _{+ 3 x + 3 x}55_{)) 11} 11

8 8

((

& 

& 

% 

% 

$$

 _"

 _"!!

##

x x  is : is : ((11))5566 ((22))6655 ((33))115544 ((44))6622 A-13.

A-13. The term containing x in the expansion ofThe term containing x in the expansion of

5 5 2 2 x x 1 1 x x

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

₍₍

 is  is --(1) 2 (1) 2ndnd (2) 3(2) 3rdrd (3) 4(3) 4thth (4) 5(4) 5thth A-14.

A-14. Given that the term of the expansion (xGiven that the term of the expansion (x1/31/3

''

 _x x''1/21/2₎₎1515 _{which does not contain x is 5 which does not contain x is 5 m, where mm, where m}

)

)

 _{N,then m= N,then m=}

((11) ) 11110000 ((22) ) 11001100 ((33) ) 11000011 ((44) ) nnoonnee

A-15.

A-15. The term independent of x in the expansion ofThe term independent of x in the expansion of

3 3 4 4 x x 1 1 x x x x 1 1 x x

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

₍₍

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

_''

 is:  is: (1) (1)

''

 3 3 ((22))00 ((33))11 ((44))33 A-16.

A-16. The term independent of x in the expansion ofThe term independent of x in the expansion of

10 10 2 2 x x 2 2 3 3 3 3 x x

!!

!!

 "

 "

 #

 #

$$

$$

% 

% 

& 

& 

((

 _is- i s-((11) ) 33//22 ((22) ) 55//44 ((33) ) 55//22 ((44) ) NNoonne e oof f tthheessee A-17.

A-17. Let the co-efficients of xLet the co-efficients of xnn _{in (1 + x) in (1 + x)}2n2n _{& (1 + x) & (1 + x)}2n2n'' 1 1 _{be P & Q respe be P & Q respe ctively, thenctively, then}

5 5 Q Q Q Q P P

_!!

 "

 "

 #

 #

$$

% 

% 

& 

&  ((

= = ((11) ) 99 ((22) ) 2277 ((33) ) 8811 ((44) ) nnoonne e oof f tthheessee A-18.

A-18. If (1 + by)If (1 + by)nn _{= (1+ 8y + 24 y = (1+ 8y + 24 y}22 _{+....) where n +....) where n}

)

)

_{N then the value of b N then the value of b and n are respectively-and n are}

respectively-((11) ) 44, , 22 ((22) ) 22,, – – 4 4 ((33) ) 22, , 44 ((44)) – – 2, 4 2, 4

A-19.

A-19. The coefficient of xThe coefficient of x5252 _{in the expansion in the expansion}

+

+

,, 100 100 0 0 m m m m 100 100_C C  (x _{(x  – –} 3) 3)100100 – –mm. 2. 2mm is  is :: (1) (1)100100_CC 47 47 (2)(2) 100 100_CC 48 48 (3)(3) – – 100 100_CC 52 52 (4)(4) – – 100 100_CC 100 100 A-20.

A-20. The co-efficient of xThe co-efficient of x55 _{in the expansion of (1 +  in the expansion of (1 + x)x)}2121 _{+ (1 + x) + (1 + x)}2222 _{+... + (1 + x) +... + (1 + x)}3030 _{is : is :}

(1) (1)5151_CC 5 5 (2)(2) 9 9_CC 5 5 (3)(3) 31 31_CC 6 6

''

21 21_CC 6 6 (4)(4) 30 30_CC 5 5 + + 20 20_CC 5 5 A-21.

A-21. The term independent of x in (1 + x)The term independent of x in (1 + x)mm

n n x x 1 1 1 1

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

₍₍

 is  is (1) (1)mm – – n nCC n n (2)(2) m + n m + n_CC n n (3)(3) m + 1 m + 1_CC n n (4)(4) m + n m + n_CC n+1 n+1 A-22.

A-22. (1 + x) (1 + x (1 + x) (1 + x + x+ x22_{) (1 + x + x) (1 + x + x}22 _{+  + xx}33_{)... (1 + x + x)... (1 + x + x}22 _{+... + x +... + x}100100_{) when written in the ascending power) when written in the ascending power}

of x then the highest exponent of x is of x then the highest exponent of x is

((11) ) 55000000 ((22) ) 55003300 ((33) ) 55005500 ((44) ) 55004400

Section (B) :

Section (B) : Numerically greatest term, Remainder and Divisibility problems

Numerically greatest term, Remainder and Divisibility problems

B-1.

B-1. The numerically greatest term in the expansion of (2 + 3The numerically greatest term in the expansion of (2 + 3 x)x)99_{, when x = 3/2 is, when x = 3/2 is}

(1) (1)99_CC

6

6. 2. 299. (3/2). (3/2)1212 (2)(2)99CC33. 2. 299. (3/2). (3/2)66 (3)(3)99CC55. 2. 299. (3/2). (3/2)1010 (4)(4)99CC44. 2. 299. (3/2). (3/2)88

B-2.

B-2. The numerically greatest term in the expansion of (2xThe numerically greatest term in the expansion of (2x ++ 5y)5y)3434_{, when x = 3 & y = 2 is :, when x = 3 & y = 2 is :}

(1) T

(1) T₂₁₂₁ (2) T(2) T₂₂₂₂ (3) T(3) T₂₃₂₃ (4) T(4) T₂₄₂₄ B-3.

B-3. The remainder when 2The remainder when 220032003 _{is divided by 17 is : is divided by 17 is :}

C-9.

C-9. The value ofThe value of

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

0 0 50 50

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

1 1 50 50  +  +

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

1 1 50 50

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

2 2 50 50  +...+  +...+

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

49 49 50 50

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

50 50 50 50  is, where  is, where nn_CC rr = =

 _"

 _"

!!!!

 #

 #

$$$$

% 

% 

& 

& 

rr n n (1) (1)

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

50 50 100 100 (2) (2)

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

51 51 100 100 (3) (3)

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

25 25 50 50 (4) (4) 2 2 25 25 50 50

!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

C-10.

C-10. The value ofThe value of

+

+

,, '' 10 10 1 1 rr nn rr 11 rr n n C C C C ..

rr  is equal to is equal to (1) 5 (2n

(1) 5 (2n – – 9) 9) ((22) ) 110 0 nn ((33) ) 9 9 ((nn – – 4) 4) (4) none of these(4) none of these

C-11.

C-11. The value of the expressionThe value of the expression

_!!

!!

 "

 "

 #

 #

$$

$$

% 

% 

& 

& 

+

+

,, 10 10 0 0 rr rr 10 10_CC

!!

!!

 "

 "

 #

 #

$$

$$

% 

% 

& 

& 

''

+

+

,, 10 10 0 0 K K K K K K 10 10 K K 2 2 C C )) 1 1 ((  _{is : is :} (1) 2 (1) 21010 _{(2) 2(2) 2}2020 _{((33))11 ((44))22}55 C-12.

C-12. In the expansion of (1 + x)In the expansion of (1 + x)nn

n n x x 1 1 1 1

!!

 "

 "

 #

 #

$$

% 

% 

& 

& 

₍₍

, the term independent of x , the term independent of x is-(1)

(1) CC  + 22₀₀2 + 2 CC  +...+ (n + 1)2₁₁2 +...+ (n + 1) CC_nn22 (2) (C(2) (C₀₀+ C+ C₁₁ +....+ C +....+ C_nn))22 (3)

(3) CC  +22₀₀ + C  +...+C₁₁22 +...+ CC22_nn (4) None of these(4) None of these C-13.

C-13. If (1 + x)If (1 + x)nn = C = C₀₀ + C + C₁₁x + Cx + C₂₂xx22 +...+C +...+C_nn.x.xnn then for n odd, C then for n odd, C₁₁22 + C + C₃₃22 + C + C₅₅22 +...+ C +...+ C_nn22 is equal to is equal to (1) 2 (1) 22n2n – – 2 2 (2) 2(2) 2nn (3)(3) ₂₂ )) !! n n (( 2 2 )! )! n n 2 2 (( (4) (4) 22 )) !! n n (( )! )! n n 2 2 (( C-14. C-14. If aIf a_nn = =

+

+

,, n n 0 0 rr rr n n_CC 1 1 , the value of , the value of

+

+

,,

''

n n 0 0 rr rr n n_CC rr 2 2 n n  is :  is : (1) (1) 2 2 n n a a_nn (2)(2) 4 4 1 1  a  a_nn (3) na(3) na_nn (4) 0(4) 0

Section (D) : Multinomial Theorem, Binomial Theorem for negative and fractional index

Section (D) : Multinomial Theorem, Binomial Theorem for negative and fractional index

D-1.

D-1. The coefficient of aThe coefficient of a55 _b b44 _c c77 _{in the expansion of (bc +  in the expansion of (bc + caca}

((

 _ab) ab)88 _is is

((11))228800 ((22))224400 ((33))118800 ((44))3322 D-2.

D-2. IfIf

4

4

xx

4

4

 < 1, then the co-efficient of x < 1, then the co-efficient of xnn _{in the expansion of (1 + x + x in the expansion of (1 + x + x}22 _{+ x + x}33 _+...) +...)22 _is is

((11))nn ((22)) nn

''

 1 1 ((33)) nn++ 22 ((44)) nn ++ 11

D-D-33 The coefficient of xThe coefficient of x44 _{in the  in the expression expression (1 + 2x + (1 + 2x + 3x3x}22 _{+ 4x + 4x}33 _{+ ...up to + ...up to}

*

*

₎₎1/21/2 _{(where | (where | x | x | < 1< 1) ) isis}

((11))11 ((22))33 ((33))22 ((44))55

Section (E) : Exponential and Logarithmic series

Section (E) : Exponential and Logarithmic series

E-1_.

E-1_. Sum of the infinite seriesSum of the infinite series !! 4 4 3 3 2 2 1 1 !! 3 3 2 2 1 1 !! 2 2 1 1

₍₍

((

₍₍

((

((

 + ... to  + ... to

*

*

(1) (1) 3 3 e e ((22))ee ((33)) 2 2 e e (4) none of these (4) none of these E-2_.

E-2_. The coefficient of xThe coefficient of x66 _{in series e in series e}2x2x _is is

(1) (1) 45 45 4 4 (2) (2) 45 45 3 3 (3) (3) 45 45 2 2 (4) none of these (4) none of these

4.

4. In the expansion ofIn the expansion of

20 20 4 4 3 3 6 6 1 1 4 4

_!!!!

 "

 "

 #

 #

$$$$

% 

% 

& 

& 

((

(1

(1) ) tthe he nnumumbeber r of of irirrarattioionanal l tetermrms s iis s 1919 (2(2) ) mimidddldle e teterm rm iis s iirrrratatiiononalal ((33) ) tthhe e nnuummbbeer r oof f rraattiioonnaal l tteerrmms s iis s 22 ((44) ) AAlll l oof f tthheessee

5. 5. If (1 + 2x + 3xIf (1 + 2x + 3x22₎₎1010 _{= a = a} 0 0 + a + a11x + x + aa22xx22 +.... + a +.... + a2020xx2020, then :, then : (1) a (1) a₁₁ = = 2200 ((22)) aa₂₂ =  = 221100 ((33) ) aa₄₄ =  = 88008855 ((44) ) AAlll l oof f tthheessee 6. 6. (1 + x + x(1 + x + x22 _{+  + xx}33₎₎55 _{=  = aa} 0 0 +  + aa11x + x + aa22xx 2 2 _{+... + a +... + a} 15 15xx 15 15_{, then a, then a} 10 10 equals to : equals to : ((11))9999 ((22))110011 ((33))110000 ((44))111100 7.

7. In the expansion ofIn the expansion of

n n 2 2 3 3 x x 1 1 x x

 _"

 _"

 #

 #

!!

$$% % 

& 

& 

_''

, n

, n

)

)

 N, if the sum of the coefficients of x N, if the sum of the coefficients of x55 _{and x and x}1010 _{is 0, then n is : is 0, then n is :}

((11) ) 2255 ((22) ) 2200 ((33) ) 1155 ((44) ) NNoonne e oof f tthheessee

8.

8. The coefficient of the term independent of x in the expansion ofThe coefficient of the term independent of x in the expansion of

10 10 2 2 1 1 3 3 1 1 3 3 2 2 x x x x 1 1 x x 1 1 x x x x 1 1 x x

!!

!!

!!

 "

 "

 #

 #

$$

$$

$$

% 

% 

& 

& 

''

''

''

((

''

((

 is  is :: ((11))7700 ((22))111122 ((33))110055 ((44))221100 9

9.. TThhe e tteerrm im in n tthhe e eexxppaannssiioon n oof (f (22xx – – 5) 5)66 which has greatest binomial coefficient is which has greatest binomial coefficient is (1) T

(1) T₃₃ (2) T(2) T₄₄ (3) T(3) T₅₅ (4) T(4) T₆₆ 10.

10. The remainder when 7The remainder when 79898 _{is divided by 5 is is divided by 5 is}

((11))44 ((22))00 ((33))22 ((44))33

11.

11. The last three digits of the number (27)The last three digits of the number (27)2727 _is is

((11))880055 ((22))330011 ((33))550033 ((44))880033 12.

12. 7799 _{+ 9 + 9}77 _{is divisible by : is divisible by :}

((11))77 ((22))2244 ((33))6644 ((44))7722 13.

13. Let f(n) = 10Let f(n) = 10nn _{+ 3.4 + 3.4}n +2n +2_{+ 5, n+ 5, n}

)

)

 _{N. The greatest value of  N. The greatest value of the integer which divides f(n) for all n the integer which divides f(n) for all n is :is :}

((11))2277 ((22))99 ((33))33 ((44))NNoonneeoofftthheessee 14.

14. Coefficient of xCoefficient of xnn'' 1 1 _{in the expansion of, (x + 3) in the expansion of, (x + 3)}nn _{+ (x + 3) + (x + 3)}nn'' 1 1 _{(x + 2) + (x + 3) (x + 2) + (x + 3)}nn'' 2 2 _{(x + 2) (x + 2)}22 _{+... + (x + 2) +... + (x + 2)}nn

is is :: (1) (1)n+1n+1_CC 2 2((33)) ((22))nn''11CC22((55)) ((33))n+1n+1CC22((55)) ((44))nnCC22(5)(5) 15.

15. The term in the expansion of (2xThe term in the expansion of (2x – – 5) 5)66 which has greatest numerical coefficient is which has greatest numerical coefficient is (1) T

(1) T₃₃,T,T₄₄ (2) T(2) T₄₄ (3) T(3) T₅₅, T, T₆₆ (4) T(4) T₆₆, T, T₇₇ 16.

16. Number of elements in set of value of r for which,Number of elements in set of value of r for which,1818_CC

rr'' 2 2 + 2. + 2.1818CCrr'' 1 1 + +1818CCrr

55

2020CC1313 is satisfied : is satisfied :

((11) ) 4 4 eelleemmeennttss ((22) ) 5 5 eelleemmeennttss ((33) ) 7 7 eelleemmeennttss ((44) ) 110 0 eelleemmeennttss

17.

17. The number of values of 'The number of values of ' rr ' satisfying the equation,' satisfying the equation,3939

C

C

_33rr''₁₁

''

3939

C

C

_rr22  = =3939

C

C

_rr22_''11

''

3939

C

C

33rr  is : is :

((11))11 ((22))22 ((33))33 ((44))44

18.

18. The sumThe sum 11 1 1 11 1 1 2 2 22 1 1 1 1 11 !

! ( ( n n

''

)) ! !

((

! ! ( ( n n

''

)) !!

((

... ! ( ! ( nn

''

)) !! is equal to : is equal to : (1) (1) 11 n n !! (2 (2 n n'' 1 1

''

 _{1 1) ) ((22))} 22 n n !! (2 (2 n n

''

 _{1 1)) ((33))} 22 n n !!(2(2 n n''11

''

_{11)) ((44) ) nnoonnee}

PART - II : COMPREHENSION

PART - II : COMPREHENSION

Comprehension # 1 Comprehension # 1

Let P be a product

Let P be a product given by given by P = (x + aP = (x + a₁₁) (x + a) (x + a₂₂) ... (x + a) ... (x + a_nn))

a anndd LLeet t SS₁₁ = a = a₁₁ + a + a₂₂ + ... + a + ... + a_nn = =

+

+

,, n n 1 1 ii ii a a _{, S, S} 2 2 ==

+

++

+

66 j j ii  j  j ii..aa ,, a a  _S S 3 3 = =

+ +

+ ++

+

66 66 j j kk ii k k  j  j ii..aa ..aa a

a and so on,and so on, then it can be shown that

then it can be shown that P = x P = xnn _{+ S + S} 1 1 x x n n  – – 1 1 + S+ S 2 2 x x n n – – 2 2 + ... + S+ ... + S n n.. 1.

1. The coefficient of xThe coefficient of x88 _{in the expression (2 + x) in the expression (2 + x)}22 _{(3 + x) (3 + x)}33 _{(4 + x) (4 + x)}44 _{must be must be}

((11))2266 ((22))2277 ((33))2288 ((44))2299 2.

2. The coefficient of xThe coefficient of x1919 _{in the expression (x in the expression (x –}

 – 1) (x 1) (x – – 2 222) (x) (x – – 3 322) ... (x) ... (x – – 20 2022) must be) must be ((11) ) 22887700 ((22) ) 22880000 ((33)) – –28702870 (4)(4) – – 4100 4100 3.

3. The coefficient of xThe coefficient of x9898 _{in the expression of (x in the expression of (x –}

 – 1) (x 1) (x – – 2) ... (x 2) ... (x – – 100) must be 100) must be (1) 1 (1) 122 _{+ 2 + 2}22 _{+ 3 + 3}22 _{+ ... + 100 + ... + 100}22 (2) (1 + 2 + 3 + ... + 100) (2) (1 + 2 + 3 + ... + 100)22  –  – (1 (122 + 2 + 222 + 3 + 322 + ... + 100 + ... + 10022)) (3) (3) 2 2 1 1 [(1 + 2 + 3 + ... + 100) [(1 + 2 + 3 + ... + 100)22  –  – (1 (122 + 2 + 222 + 3 + 322 + ... + 100 + ... + 10022)])] (4)

(4) None None of theseof these Comprehension Comprehension # # 22 We know that if We know that if nn_CC 0 0,, n n_CC 1 1,, n n_CC 2 2, ...,, ..., n n_CC n

n be binomial coefficients, then (1 + x) be binomial coefficients, then (1 + x) n n _{= C = C} 0 0 + C + C11 x + C x + C22 x x 2 2 _{+ C + C} 3 3xx 3 3 + ...+ C

+ ...+ C_nn xxnn_{. Various relations among binomial coefficients can be derived by putting. Various relations among binomial coefficients can be derived by putting}

x = 1, x = 1,  – – 1, i, 1, i,

7

7

_!!

!!

 "

 "

 #

 #

$$

$$

% 

% 

& 

& 

((

''

,,

7

7

''

,,

2 2 3 3 ii 2 2 1 1 ,, 1 1 ii where where .. 4.

4. The value ofThe value ofnn_CC 0 0 – – n n_CC 2 2 + + n n_CC 4 4 – – n n_CC 6 6 + ... must be + ... must be

((11))22ii ((22))((11 – – i) i)nn – – (1 + i) (1 + i)nn

(3) (3) 2 2 1 1 [(1 [(1 – – i) i)nn + (1 + i)+ (1 + i)nn]] ((44)) 2 2 1 1 [(2

[(2 – – i) i)nn + (1+ (1 – – i) i)nn]]

5.

5. The value of expression (The value of expression (nn_CC 0 0 – – n n_CC 2 2 + + n n_CC 4 4 – – n n_CC 6 6 + ...) + ...) 2 2 _{+ ( + (}nn_CC 1 1 – – n n_CC 3 3 + + n n_CC 5 5 ...) ...) 2 2 _{must be must be} (1) 2

(1) 22n2n _{(2) 2(2) 2}nn _{(3)(3) 22}nn22 _{(4) None of these(4) None of these}

PART -

PART - I :

I : AIEEE PROBLEMS

AIEEE PROBLEMS (LAST 10

(LAST 10 YEARS)

YEARS)

1.

1. IfIf nnCC   denotes the number of combinations of n things taken r at a time, then the expression_rr  denotes the number of combinations of n things taken r at a time, then the expression

rr n n 1 1 rr n n 1 1 rr n n_CC

₍₍

_CC

₍₍

₂₂

₈₈

_CC ''

((  equals equals [AIEEE 2003][AIEEE 2003]

(1)

(1) nn((22CC_rr (2)(2) nn((22CC_rr((11 (3)(3) nn((11CC_rr (4)(4) nn((11CC_rr((11

2.

2. The number of integral terms in the expansion ofThe number of integral terms in the expansion of

9 9

₃₃

 ₍

 ₍

88₅₅

::

256256 _{is  is :: [AIEEE 2003][AIEEE 2003]}