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Bino mial ex pressi on :

Any algebraic expression which contains two dissimilar terms is called binomial expression. For example : x + y, x2_{y +} 2 xy 1 , 3 – x, x21 + _(x3 ₁₎1/3 1  etc.

Terminology used in binomial theorem :

Factorial notation : or n! is pronounced as factorial n and is defined as

n! =        0 n if ; 1 N n if ; 1 . 2 . 3 )... 2 n )( 1 n ( n Note : n! = n . (n – 1)! ; n 

N

Mathematical meaning of nC_r : The term nCr denotes number of combinations of r things choosen

from n distinct things mathematically, n_C

r = _{(n r)!r!}

! n



, n N, r  W, 0 r n

Note : Other symbols of of n_C

r are        r n and C(n, r).

Pr oper ties re late d to nC_r :

(i) n_C r = n_C n – r Note : If n_C x = n_C y  Either x = y or x + y = n (ii) n_C r + n_C r – 1 = n + 1_C r (iii) 1 r n r n C C  = n_rr1 (iv) n_C r = _r n n–1_C r–1 = _{r(r 1)} ) 1 n ( n   n–2_C r–2 = ... = r(r 1)(r 2)...2.1 )) 1 r ( n ( )... 2 n )( 1 n ( n      

(v) If n and r are relatively prime, then n_C

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

Statement of binomial theorem :

(a + b)n ₌ n_C 0 a n_b0 ₊ n_C 1 a n–1 _b1 ₊ n_C 2 a n–2 _b2 _+...+ n_C r a n–r _br _{+... +} n_C n a 0 _bn where n  N or (a + b)n ₌

_

  n 0 r r r n r n b a C

Note : If we put a = 1 and b = x in the above binomial expansion, then or (1 + x)n₌ n_C 0 + n_C 1 x + n_C 2 x 2 _{+... +} n_C r x r _+...+ n_C n x n or (1 + x)n₌

_

 n 0 r r r n_{C x}

Binomial Theorem

Regarding Pascal’s Triangle, we note the following :

(a) Each row of the triangle begins with 1 and ends with 1.

(b) Any entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right.

Example # 3 : The number of dissimilar terms in the expansion of (1 – 3x + 3x2– x3)20 is

(A) 21 (B) 31 (C) 41 (D) 61

Solution : (1 – 3x + 3x2– x3)20 = [(1 – x)3]20 = (1 – x)60

Therefore number of dissimilar terms in the expansion of (1 – 3x + 3x2– x3)20 is 61.

General term :

(x + y)n ₌ n_C 0 x n _y0 ₊ n_C 1 x n–1 _y1 _{+ ...+} n_C r x n–r _yr _{+ ...+} n_C n x 0 _yn

(r + 1)th _{term is called general term and denoted by T} r+1.

T_r+1 = n_C r x

n–r _yr

Note : The rth _{term from the end is equal to the (n}

– r + 2)th term from the begining, i.e. nCn – r + 1 x

r – 1 _yn – r + 1

Example # 4 : Find (i) 28th _{term of (5x + 8y)}30 _{(ii) 7}th _{term of}

9 2x 5 5 4x        Solution : (i) T_{27 + 1} = 30_C 27 (5x) 30– 27 _(8y)27 ₌ ! 27 ! 3 ! 30 (5x)3 _{. (8y)}27 (ii) 7th term of 9 x 2 5 5 x 4        T_{6 + 1} = 9_C 6 6 9 5 x 4        6 x 2 5        = _3!9₆! _! 3 5 x 4       6 x 2 5       = 3 x 10500

Example # 5 : Find the number of rational terms in the expansion of (91/4 _{+ 8}1/6₎1000_.

Solution : The general term in the expansion of

_

1/4 1/6

_

1000

8 9  is T_r+1 = 1000_C r r 1000 4 1 9          r 6 1 8         = 1000_C r 2 r 1000 3  2 r 2

The above term will be rational if exponent of 3 and 2 are integers It means 2 r 1000 and 2 r must be integers

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

Mi ddl e t erm (s) :

(a) If n is even, there is only one middle term, which is

th 2 2 n        term.

(b) If n is odd, there are two middle terms, which are

th 2 1 n        and th 1 2 1 n         terms.

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

(i) 14 2 2 x 1          _(ii) 9 3 6 a a 3         

Case -  When b a 1 1 n  

is an integer (say m), then

(i) T_r+1 > T_r when r < m (r = 1, 2, 3 ...., m – 1) i.e. T₂ > T₁, T₃ > T₂, ..., T_m > T_m–1 (ii) T_r+1 = T_r when r = m i.e. T_m+1 = T_m (iii) T_r+1 < T_r when r > m (r = m + 1, m + 2, ...n ) i.e. T_m+2 < T_m+1 , T_m+3 < T_m+2 , ...T_n+1 < T_n Conclusion : When b a 1 1 n  

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

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

(i) T_r+1 > T_r when r < b a 1 1 n   (r = 1, 2, 3,..., m–1, m) i.e. T₂ > T₁ , T₃ > T₂, ..., T_m+1 > T_m (ii) T_r+1 < T_r when r > b a 1 1 n   (r = m + 1, m + 2, ...n) i.e. T_m+2 < T_m+1 , T_m+3 < T_m+2 , ..., T_{n +1} < T_n Conclusion : When b a 1 1 n  

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

term.

Note : (i) In any binomial expansion, the middle term(s) has greatest binomial coefficient. In the expansion of (a + b)n

If n No. of greatest binomial coefficient Greatest binomial coefficient

Even 1 n_C n/2 Odd 2 n_C (n – 1)/2 and n_C (n + 1)/2

(Values of both these coefficients are equal ) (ii) In order to obtain the term having numerically greatest coefficient, put a = b = 1, and proceed

as discussed above.

Example # 8 : Find the numerically greatest term in the expansion of (3 – 5x)15 when x =

5 1 . Solution : Let rth _{and (r + 1)}th _{be two consecutive terms in the expansion of (3}

– 5x)15 T_{r + 1} T_r 15_C r 3 15 – r _(| – 5x|)r15C_{r – 1} 315 – (r – 1) (|– 5x|)r – 1 ! r ! ) r 15 ( )! 15   |– 5x |  ! ) 1 r ( ! ) r 16 ( )! 15 . 3    5 . 5 1 (16 – r) 3r 16 – r  3r 4r  16 r  4

Self practice problems :

(8) If n is a positive integer, then show that 32n + 1 _{+ 2}n + 2 _{is divisible by 7.}

(9) What is the remainder when 7103 _{is divided by 25 .}

(10) Find the last digit, last two digits and last three digits of the number (81)25.

(11) Which number is larger (1.2)4000 _{or 800}

Answers : (9) 18 (10) 1, 01, 001 (11) (1.2)4000_.

Some standard expansi ons :

(i) Consider the expansion

(x + y)n ₌

_

 n 0 r r n C xn–r _yr ₌ n_C 0 x n _y0 ₊ n_C 1 x n–1 _y1 _{+ ...+} n_C r x n–r _yr _{+ ...+} n_C n x 0 _yn _....(i)

(ii) Now replace y – y we get

(x – y)n =



 n 0 r r n C (– 1) r xn–r yr = nC0 x n _y0 – nC1 x n–1 _y1 _{+ ...+} n_C r (–1) r _xn–r _yr _{+ ...+} n_C n (– 1) n _x0 _yn _....(ii)

(iii) Adding (i) & (ii), we get (x + y)n _{+ (x}

– y)n = 2[nC0 x

n _y0 ₊ n_C 2 x

n – 2 _y2 _+...]

(iv) Subtracting (ii) from (i), we get (x + y)n

– (x – y)n = 2[nC1 x

n – 1 _y1 ₊ n_C 3 x

n – 3 _y3 _+...]

Properties of binomial coefficients :

(1 + x)n _{= C} 0 + C1x + C2x 2 _{+ ... + C} r x r _{+ ... + C} nx n _...(1) where C_r denotes n_C r

(1) The sum of the binomial coefficients in the expansion of (1 + x)n _{is 2}n

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

_

  n 0 r n r n 2 C

(2) Again putting x = –1 in (1), we get n_C 0– n_C 1 + n_C 2– n_C 3 + ... + (–1) nn_C n = 0 ...(3) or

_

   n 0 r r n r _{C 0} ) 1 (

(3) The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficients at even position and each is equal to 2n–1_.

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

(4) Sum of two consecutive binomial coefficients

n_C r + n_C r–1 = n+1_C r  L.H.S. = n_C r + n_C r–1 = _{(n r)!r!} ! n  + (n r 1)!(r 1)! ! n    = _{(n r)!}n!_{(r 1)!}           1 r n 1 r 1 = _{(n r)!}n!_{(r 1)!}   r(n r 1) ) 1 n (    = _(n(n_r 1₁)!_)!r!    = n+1_C r = R.H.S.

Method : By Integration (1 + x)n _{= C} 0 + C1x + C2x 2 _{+ ... + C} n x n_.

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

0 1 1 n 1 n ) x 1 (             = 0 1 1 n n 3 2 2 1 0 1 n x C ... 3 x C 2 x C x C                1 n 1  – 0 = 0 –              1 n C ) 1 ( ... 3 C 2 C C₀ 1 2 n 1 n C₀– 2 C1 + 3 C2 – ... + (– 1)n 1 n Cn  = n 1 1  Proved Example # 14 : If (1 + x)n _{= C} 0 + C1x + C2x 2 _{+ ...+ C} nx

n_{, then prove that}

(i) C₀2 _{+ C} 1 2 _{+ C} 2 2 _{+ ... + C} n 2 ₌ 2n_C n (ii) C₀C₂ + C₁C₃ + C₂C₄ + ... + C_{n – 2} C_n = 2n_C n – 2 or 2n_C n + 2 (iii) 1. C₀2 _{+ 3 . C} 1 2 _{+ 5. C} 2 2 _{+ ... + (2n + 1) . C} n 2 _{. = 2n.} 2n – 1_C n + 2n_C n. Solution : (i) (1 + x)n _{= C} 0 + C1x + C2x 2 _{+ ... + C} n x n_{. ...(i)} (x + 1)n _{= C} 0x n _{+ C} 1x n – 1_{+ C} 2x n – 2 _{+ ... + C} n x 0 _...(ii)

Multiplying (i) and (ii)

(C₀ + C₁x + C₂x2 _{+ ... + C} nx n_{) (C} 0x n _{+ C} 1x n – 1 _{+ ... + C} nx 0_{) = (1 + x)}2n Comparing coefficient of xn_, C₀2 _{+ C} 1 2 _{+ C} 2 2 _{+ ... + C} n 2 ₌ 2n_C n

(ii) From the product of (i) and (ii) comparing coefficients of xn – 2 _{or x}n + 2 _{both sides,}

C₀C₂ + C₁C₃ + C₂C₄ + ... + C_{n – 2}C_n = 2n_C n – 2or

2n_C n + 2.

(iii)  Method : By Summation

L.H.S. = 1. C₀2 _{+ 3. C} 1 2 _{+ 5. C} 2 2 _{+ ... + (2n + 1) C} n 2_. =

_

  n 0 r ) 1 r 2 ( n_C r 2 ₌



 n 0 r r . 2 . (n_C r) 2 ₊



 n 0 r 2 r n ) C ( = 2

_

 n 1 r n . . n – 1_C r – 1 n_C r + 2n_C n (1 + x)n ₌ n_C 0 + n_C 1 x + n_C 2 x 2 _{+ ...}n_C n x n _...(i) (x + 1)n – 1 ₌ n – 1_C 0 x n – 1 ₊ n – 1_C 1 x n – 2 _{+ ...+}n – 1_C n – 1x 0 _...(ii)

Multiplying (i) and (ii) and comparing coeffcients of xn_. n – 1_C 0 . n_C 1 + n – 1_C 1 . n_C 2 + ... + n – 1_C n – 1. n_C n = 2n – 1_C n



   n 0 r 1 r 1 n C . n_C r = 2n – 1_C n

Hence, required summation is 2n. 2n – 1_C n + 2n_C n = R.H.S.  Method : By Differentiation (1 + x2₎n _{= C} 0 + C1x 2 _{+ C} 2x 4 _{+ C} 3x 6 _{+ ...+ C} n x 2n

Multiplying both sides by x x(1 + x2₎n _{= C} 0x + C1x 3 _{+ C} 2x 5 _{+ ... + C} nx 2n + 1_.

Differentiating both sides

x . n (1 + x2₎n – 1 _{. 2x + (1 + x}2₎n _{= C} 0 + 3. C1x 2 _{+ 5. C} 2 x 4 _{+ ...+ (2n + 1) C} n x 2n_...(i) (x2 _{+ 1)}n _{= C} 0 x 2n _{+ C} 1 x 2n – 2 _{+ C} 2 x 2n – 4 _{+ ... + C} n ...(ii)

Multiplying (i) & (ii) (C₀ + 3C₁x2 _{+ 5C} 2x 4 _{+ ... + (2n + 1) C} n x 2n_{) (C} 0 x 2n _{+ C} 1x 2n – 2 _{+ ... + C} n) = 2n x2 _{(1 + x}2₎2n – 1 _{+ (1 + x}2₎2n comparing coefficient of x2n_, C₀2 _{+ 3C} 1 2 _{+ 5C} 2 2 _{+ ...+ (2n + 1) C} n 2_{= 2n .} 2n – 1_C n – 1 + 2n_C n. C₀2 _{+ 3C} 1 2 _{+ 5C} 2 2 _{+ ...+ (2n + 1) C} n 2_{= 2n .} 2n–1_C n + 2n_C n. Proved

Mul tino mial the orem :

As we know the Binomial Theorem –

(x + y)n ₌



 n 0 r r n C xn–r _yr ₌



  n 0 r (n r)!r! ! n xn–r _yr putting n – r = r₁ , r = r₂ therefore, (x + y)n ₌

_

 r n r_{1 2} r1! r2! ! n _xr1_.yr2

Total number of terms in the expansion of (x + y)n _{is equal to number of non-negative integral solution}

of r₁ + r₂ = n i.e. n+2–1_C 2–1 =

n+1_C

1 = n + 1

In the same fashion we can write the multinomial theorem (x₁ + x₂ + x₃ + ... x_k)n ₌



   r ... r n r_{1 2 k} r1!r2!...rk! ! n _{1 2} r_k k r 2 r 1 .x ...x x

Here total number of terms in the expansion of (x₁ + x₂ + ... + x_k)n _{is equal to number of}

non-negative integral solution of r₁ + r₂ + ... + r_k = n i.e. n+k–1_C k–1

Example # 17 : Find the coefficient of a2 _b3 _c4 _{d in the expansion of (a}

– b – c + d)10 Solution : (a – b – c + d)10 =



   r r r 10 r_{1 2 3 4} r1!r2!r3!r4! )! 10 ( 4 3 2 1 r r r r ) d ( ) c ( ) b ( ) a (  

we want to get a2 _b3 _c4 _{d this implies that r}

1 = 2, r2 = 3, r3 = 4, r4 = 1  coeff. of a2 b3 c4 d is _2!3!4!1! )! 10 ( (–1)3 (–1)4 = – 12600

Example # 18 : In the expansion of

11 x 7 x 1       

 , find the term independent of x.

Solution : 11 x 7 x 1         =



  r r 11 r_{1 2 3} r1!r2!r3! )! 11 ( 3 2 1 r r r x 7 ) x ( ) 1 (      

The exponent 11 is to be divided among the base variables 1, x and x 7

in such a way so that we get x0_.

Therefore, possible set of values of (r₁, r₂, r₃) are (11, 0, 0), (9, 1, 1), (7, 2, 2), (5, 3, 3), (3, 4, 4), (1, 5, 5)

Hence the required term is

)! 11 ( )! 11 ( (70_{) +} ! 1 ! 1 ! 9 )! 11 ( 71 ₊ ! 2 ! 2 ! 7 )! 11 ( 72 ₊ ! 3 ! 3 ! 5 )! 11 ( 73 ₊ ! 4 ! 4 ! 3 )! 11 ( 74 ₊ ! 5 ! 5 ! 1 )! 11 ( 75 = 1 + ! 2 ! 9 )! 11 ( . ! 1 ! 1 ! 2 71 ₊ ! 4 ! 7 )! 11 ( . ! 2 ! 2 ! 4 72 ₊ ! 6 ! 5 ! ) 11 ( . ! 3 ! 3 ! 6 73 + ! 8 ! 3 ! ) 11 ( . ! 4 ! 4 ! 8 74 ₊ ! 10 ! 1 ! ) 11 ( . ! 5 ! 5 ! ) 10 ( 75 = 1 + 11_C 2 . 2_C 1 . 7 1 ₊ 11_C 4 . 4_C 2 . 7 2 ₊ 11_C 6 . 6_C 3 . 7 3 ₊ 11_C 8 . 8_C 4 . 7 4 ₊ 11_C 10 . 10_C 5 . 7 5 = 1 +

_

 5 1 r r 2 11_C . 2r_C r . 7 r

Example-20 : If x is so small such that its square and higher powers may be neglected, then find the value of 2 / 1 3 / 5 2 / 1 ) x 4 ( ) x 1 ( ) x 3 1 (     Solution : _1/2 3 / 5 2 / 1 ) x 4 ( ) x 1 ( ) x 3 1 (     = 1/2 4 x 1 2 3 x 5 1 x 2 3 1           = 2 1        x 6 19 2 2 / 1 4 x 1         = 2 1        x 6 19 2        8 x 1 ₌ 2 1         x 6 19 4 x 2 _{= 1 –} 8 x – 12 19 x = 1 – 24 41 x Self practice problems :

(16) Find the possible set of values of x for which expansion of (3 – 2x)1/2 is valid in ascending

powers of x. (17) If y = 5 2 + 1₂.3_! 2 5 2       + ! 3 5 . 3 . 1 3 5 2      

+ ..., then find the value of y2 _{+ 2y}

(18) The coefficient of x100 _in 2 ) x 1 ( x 5 3   is (A) 100 (B) –57 (C) –197 (D) 53 Answers : (16) x         2 3 , 2 3 (17) 4 (18) C

20. Find the coefficient of the term independent of x in the expansion of 10 2 / 1 3 / 1 3 / 2 _{x x} 1 x 1 x x 1 x             21. If in the expansion of (1 + x)m ₍₁

– x)n, the coefficients of x and x2 are 3 and –6 respectively. Then find the value

of m.

22. _{Find the number of terms in the expansion of (1 + 5 2 x)}9 _{+ (1}

– 5 2 x)9.

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

of n.

24. If in the expansion of (1 – x)2n–1 the coefficient of xr is denoted by ar, then prove that ar–1 + a2n–r = 0

25. Using binomial theorem, prove that 23n

– 7n – 1 is divisible by 49 where n  N.

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

– 8n – 9 is divisible by 64, n  N. 27. Prove that 5 . 4 1 – 4 . 3 1 3 . 2 1 – 2 . 1 1  + ...  = log_e       e 4 .

28. Find the sum of the infinite series 1 +

! 6 1 ! 4 1 ! 2 1   _{+ ...} 29. Prove that (x2– y2) + _2! 1 (x4– y4) + _3! 1 (x6– y6) + ... to  = 2 2 y x _e – e

Type (IV) : Very Long Answer Type Questions: [06 Mark Each]

30. Find the value of

_

    n 0 r r n e e r n r ) 10 log 1 ( 10 log r 1 C ) 1 ( .

31. If the coefficient of rth_{, (r + 1)}th_{and (r + 2)}th _{terms in the expansion of (1 + x)}14 _{are in A.P, then find}

the value of r.

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

33. If 3rd_{, 4}th_{, 5}th _{and 6}th _{terms in the expansion of (x +}

)n be respectively a, b, c and d then prove that

bd c ac b 2 2   = c 3 a 5

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

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

36. Sum the series from n = 1 to n =  , whose nth term is

(i) ! ) 1 n ( 1  (ii) (n 2)! 1  (iii) (2n–1)! 1 37. Prove that log_e       n m = 2                                 .... n m n – m 5 1 n m n – m 3 1 n m n – m 3 5 38. Prove that log_e        x 1 x =             .... ) 1 x 2 ( 5 1 ) 1 x 2 ( 3 1 ) 1 x 2 ( 1 2 _{3 5}

A-12. The co-efficient of x in the expansion of (1  2x3 + 3 x5)

1

1

8















x

is : (1) 56 (2) 65 (3) 154 (4) 62

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

5 2 x 1 x        is -(1) 2nd _{(2) 3}rd _{(3) 4}th _{(4) 5}th

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

 x1/2)15 which does not contain x is 5m, where m N,then m=

(1) 1100 (2) 1010 (3) 1001 (4) none

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

3 4 x 1 x x 1 x               is: (1)  3 (2) 0 (3) 1 (4) 3

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

10 2 x 2 3 3 x          is-(1) 3/2 (2) 5/4 (3) 5/2 (4) None of these

A-17. Let the co-efficients of xn _{in (1 + x)}2n _{& (1 + x)}2n  1 _{be P & Q respectively, then}

5 Q Q P        ₌ (1) 9 (2) 27 (3) 81 (4) none of these

A-18. If (1 + by)n _{= (1+ 8y + 24 y}2 _{+....) where n}

N then the value of b and n are

respectively-(1) 4, 2 (2) 2, – 4 (3) 2, 4 (4) – 2, 4

A-19. The coefficient of x52 _{in the expansion}



 100 0 m m 100 C _{(x – 3)}100–m_{. 2}m _{is :} (1) 100_C 47 (2) 100_C 48 (3) – 100_C 52 (4) – 100_C 100

A-20. The co-efficient of x5 _{in the expansion of (1 + x)}21 _{+ (1 + x)}22 _{+... + (1 + x)}30 _{is :}

(1) 51_C 5 (2) 9_C 5 (3) 31_C 6 21_C 6 (4) 30_C 5 + 20_C 5

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

n x 1 1        is (1) m – n_C n (2) m + n_C n (3) m + 1_C n (4) m + n_C n+1

A-22. (1 + x) (1 + x + x2_{) (1 + x + x}2 _{+ x}3_{)... (1 + x + x}2 _{+... + x}100_{) when written in the ascending power}

of x then the highest exponent of x is

(1) 5000 (2) 5030 (3) 5050 (4) 5040

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

B-1. The numerically greatest term in the expansion of (2 + 3x)9_{, when x = 3/2 is}

(1) 9_C

6. 29. (3/2)12 (2) 9C3. 29. (3/2)6 (3) 9C5. 29. (3/2)10 (4) 9C4. 29. (3/2)8

B-2. The numerically greatest term in the expansion of (2x+5y)34_{, when x = 3 & y = 2 is :}

(1) T₂₁ (2) T₂₂ (3) T₂₃ (4) T₂₄

B-3. The remainder when 22003 _{is divided by 17 is :}

C-9. The value of _       0 50         1 50 + _       1 50         2 50 +...+ _       49 50         50 50 is, where n_C r =        r n (1) _       50 100 (2) _       51 100 (3) _       25 50 (4) 2 25 50         C-10. The value of

_

  10 1 r r 1 n r n C C . r is equal to (1) 5 (2n – 9) (2) 10 n (3) 9 (n – 4) (4) none of these

C-11. The value of the expression

       



 10 0 r r 10 C         



 10 0 K K K 10 K 2 C ) 1 ( is : (1) 210 _{(2) 2}20 _{(3) 1 (4) 2}5 C-12. In the expansion of (1 + x)n n x 1 1      

 , the term independent of x is-(1) C2₀ + 2 C₁2 +...+ (n + 1) C_n2 (2) (C₀ + C₁ +....+ C_n)2 (3) C2₀ + C2₁ +...+ C2_n (4) None of these

C-13. If (1 + x)n = C₀ + C₁x + C₂x2 +...+C_n.xn then for n odd, C₁2 + C₃2 + C₅2 +...+ C_n2 is equal to

(1) 22n – 2 _{(2) 2}n ₍₃₎ 2 ) ! n ( 2 )! n 2 ( (4) _(n!)2 )! n 2 ( C-14. If a_n =

_

 n 0 r nCr 1 , the value of

_

  n 0 r nCr r 2 n is : (1) 2 n a_n (2) 4 1 a_n (3) na_n (4) 0

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

D-1. The coefficient of a5 _b4 _c7 _{in the expansion of (bc + ca  ab)}8 _is

(1) 280 (2) 240 (3) 180 (4) 32

D-2. If x < 1, then the co-efficient of xn in the expansion of (1 + x + x2 + x3 +...)2 is

(1) n (2) n  1 (3) n + 2 (4) n + 1

D-3 The coefficient of x4 _{in the expression (1 + 2x + 3x}2 _{+ 4x}3 _{+ ...up to}

)1/2 (where | x | < 1) is

(1) 1 (2) 3 (3) 2 (4) 5

Section (E) : Exponential and Logarithmic series

E-1_. Sum of the infinite series ! 4 3 2 1 ! 3 2 1 ! 2 1      + ... to  (1) 3 e (2) e (3) 2 e (4) none of these

E-2_. The coefficient of x6 _{in series e}2x _is

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

4. In the expansion of 20 4 3 6 1 4         

(1) the number of irrational terms is 19 (2) middle term is irrational (3) the number of rational terms is 2 (4) All of these

5. If (1 + 2x + 3x2₎10 _{= a} 0 + a1x + a2x2 +.... + a20x20, then : (1) a₁ = 20 (2) a₂ = 210 (3) a₄ = 8085 (4) All of these 6. (1 + x + x2 _{+ x}3₎5 _{= a} 0 + a1x + a2x 2 _{+... + a} 15x 15_{, then a} 10 equals to : (1) 99 (2) 101 (3) 100 (4) 110 7. In the expansion of n 2 3 x 1 x _     

 , n  N, if the sum of the coefficients of x5 and x10 is 0, then n is :

(1) 25 (2) 20 (3) 15 (4) None of these

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

10 2 1 3 1 3 2 x x 1 x 1 x x 1 x                 is : (1) 70 (2) 112 (3) 105 (4) 210

9. The term in the expansion of (2x – 5)6 which has greatest binomial coefficient is

(1) T₃ (2) T₄ (3) T₅ (4) T₆

10. The remainder when 798 _{is divided by 5 is}

(1) 4 (2) 0 (3) 2 (4) 3

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

(1) 805 (2) 301 (3) 503 (4) 803

12. 79 _{+ 9}7 _{is divisible by :}

(1) 7 (2) 24 (3) 64 (4) 72

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

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

(1) 27 (2) 9 (3) 3 (4) None of these

14. Coefficient of xn  1 _{in the expansion of, (x + 3)}n _{+ (x + 3)}n  1 _{(x + 2) + (x + 3)}n  2 _{(x + 2)}2 _{+... + (x + 2)}n

is : (1) n+1_C

2(3) (2) n1C2(5) (3) n+1C2(5) (4) nC2(5)

15. The term in the expansion of (2x – 5)6 which has greatest numerical coefficient is

(1) T₃ ,T₄ (2) T₄ (3) T₅ , T₆ (4) T₆ , T₇

16. Number of elements in set of value of r for which, 18_C

r  2 + 2.

18_C r  1 +

18_C

r20C13 is satisfied :

(1) 4 elements (2) 5 elements (3) 7 elements (4) 10 elements

17. The number of values of 'r' satisfying the equation,39

C

_3r1



39

C

_r2 =39

C

_r2 ₁



39

C

3r

 is : (1) 1 (2) 2 (3) 3 (4) 4 18. The sum

1

1

1

1

2

2

1

1

1

! (

n



) !

! (

n

) !

...

! (

n

) !









is equal to : (1)

1

n !

(2 n  1  1) (2)

2

n !

(2 n  1) (3)

2

n !

(2 n1 1) (4) none

PART - II : COMPREHENSION

Comprehension # 1

Let P be a product given by P = (x + a₁) (x + a₂) ... (x + a_n) and Let S₁ = a₁ + a₂ + ... + a_n =

_

 n 1 i i a , S₂ =

_{ }

j i j i.a , a S₃ =

_{ }

 j k i k j i.a.a a and so on, then it can be shown that

P = xn _{+ S} 1 x n – 1 _{+ S} 2 x n – 2 _{+ ... + S} n.

1. The coefficient of x8 _{in the expression (2 + x)}2 _{(3 + x)}3 _{(4 + x)}4 _{must be}

(1) 26 (2) 27 (3) 28 (4) 29

2. The coefficient of x19 _{in the expression (x}

– 1) (x – 22) (x – 32) ... (x – 202) must be

(1) 2870 (2) 2800 (3) –2870 (4) – 4100

3. The coefficient of x98 _{in the expression of (x}

– 1) (x – 2) ... (x – 100) must be (1) 12 _{+ 2}2 _{+ 3}2 _{+ ... + 100}2 (2) (1 + 2 + 3 + ... + 100)2 – (12 + 22 + 32 + ... + 1002) (3) 2 1 [(1 + 2 + 3 + ... + 100)2 – (12 + 22 + 32 + ... + 1002)] (4) None of these Comprehension # 2 We know that if n_C 0, n_C 1, n_C 2, ..., n_C

n be binomial coefficients, then (1 + x) n _{= C}

0 + C1 x + C2 x 2 _{+ C}

3x 3

+ ...+ C_n xn_{. Various relations among binom ial coefficients can be derived by putting}

x = 1, – 1, i,                2 3 i 2 1 , 1 i where . 4. The value of n_C 0– n_C 2 + n_C 4– n_C 6 + ... must be (1) 2i (2) (1 – i)n – (1 + i)n (3) 2 1 [(1 – i)n + (1 + i)n] (4) 2 1 [(2 – i)n + (1 – i)n]

5. The value of expression (n_C 0– n_C 2 + n_C 4– n_C 6 + ...) 2 _{+ (}n_C 1– n_C 3 + n_C 5 ...) 2 _{must be} (1) 22n _{(2) 2}n ₍₃₎ n2 2 (4) None of these

PART - I : AIEEE PROBLEMS

(LAST 10 YEARS)

1. If nC_r denotes the number of combinations of n things taken r at a time, then the expression

r n 1 r n 1 r n_{C C 2 C}    _  equals [AIEEE 2003] (1) n2_{Cr (2)} 1 r 2 n _C   ₍₃₎ r 1 n _{C (4)} 1 r 1 n _C  

2. The number of integral terms in the expansion of

_

₃ 8₅

_

256

 is : [AIEEE 2003]

13. The sum of the series 20_C 0– 20_C 1 + 20_C 2– 20_C 3 + ... + 20_C 10 is [AIEEE 2007 (3, –1), 120] (1) –20C10 (2) ₂ 1 ₂₀ C₁₀ (3) 0 (4) 20_C 10

14. In the binomial expansion of (a – b)n , n  5, the sum of 5th and 6th term is zero, then

b a equals [AIEEE 2008 (3, –1), 105] (1) 5 4 n (2) 4 n 5  (3) 5 n 6  (4) 6 5 n 15. Statement-1 :



  n 0 r ) 1 r ( nC_r = (n + 2) 2n–1 _{[AIEEE 2008 (3,} –1), 105] Statement-2 :

_

 n 0 r (r + 1) n_C r xr = (1 + x)n + nx (1 + x)n – 1

(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True

16. Let S₁ =

_

 10 1 j ) 1 – j ( j 10_C j , S2 =



 10 1 j j10_C j and S3 =



 10 1 j 2 j 10_C j. [AIEEE 2009 (4, –1), 144] Statement -1 : S₃ = 55 × 29 . Statement -2 : S₁ = 90 × 28 and S₂ = 10 × 28.

(1) Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement -1. (2) Statement-1 is true, Statement-2 is false.

(3) Statement -1 is false, Statement -2 is true.

(4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.

17. The coefficient of x7 _{in the expansion of (1}

– x – x2 + x3)6 is : [AIEEE 2011 (4, –1), 120]

(1) 144 (2) – 132 (3) – 144 (4) 132

18. If n is a positive integer, then

_

_31

_

2n –



31



2n is : [AIEEE-2012, (4, –1)/120]

(1) an irrational number (2) an odd positive integer

(3) an even positive integer (4) a rational number other than positive integers

19. The term independent of x in expansion of

10 2 / 1 3 / 1 3 / 2 x x 1 x 1 x x 1 x             is : [AIEEE - 2013, (4, – ¼) 120 ] (1) 4 (2) 120 (3) 210 (4) 310

BOARD LEVEL SOLUTIONS

Type (I) 1.

 

º 4 3 0 4 x 4 x C       + 1 3 3 1 4 x 4 ) x ( C       + 2 2 3 2 4 x 4 ) x ( C       + 3 1 3 3 4 x 4 ) x ( C       + 4 0 3 4 4 x 4 ) x ( C        x12 + 16 x8 + 96 x4 + 256 + 4 x 256 Ans. 2. 6_C0(ax)6 º x b        + 6C₁ (ax)5 1 x b –       + 6_C 2 (ax) 4 2 x b –       + 6_C 3 (ax) 3 3 x b –       + 6_C 4 (ax) 2 4 x b –       + 6_C 5 ax 5 x b –       + 6_C 6 (ax) 6 6 x b –       = a6_x6 – 6a5bx4 + 15a4b2x2– 20a3b3 + ₂ 4 2 x b a 15 – ₄ 5 x ab 6 + ₆ 6 x b 3. 0 4 0 4 x a a x C                  + 1 3 1 4 x a a x C                  + 2 2 2 4 x a a x C                  + 3 1 3 4 x a a x C                  + 4 0 4 4 x a a x C                  = ₂ 2 a x – a x 4 + 6 – 4 x a + ₂ 2 x a Ans. 4.  e2x+3 = e2x.e3 = e3             ... ! 3 ) x 2 ( ! 2 ) x 2 ( ! 1 ) x 2 ( 1 3 2

Thus the coefficient of x2 in the expansion of e2x+3 _{is e}32 2 = 2e3 5. We have, ex _{= 1 +} ! 4 x ! 3 x ! 2 x ! 1 x 2 3 4    + .... to  Putting x = 2, we get e2 _{= 1 +} ! 7 2 ! 6 2 ! 5 2 ! 4 2 ! 3 2 ! 2 2 ! 1 2 2 3 4 5 6 7       +...  e2 = 1 + 2 + 2 + 1·333 + 0.666 + 0.266 + 0.088 + 0.025  e2 = 7·378

e2 = 7·4 (correct to one decimal place)

6. log_e(1 + 3x + 2x2_{) = log} e[(1 + 2x) (1 + x)] = log_e(1 + 2x) + log_e(1 + x) =            (2x) ... 4 1 – ) x 2 ( 3 1 2 ) x 2 ( – x 2 3 4 2 + _         (x) ... 4 1 – ) x ( 3 1 ) x ( 2 1 – x 2 3 4 = 3x– 2 5 x2 _+3x3 – 4 17 x4 _{+ ...}  Type (II) 7.  5C₀(1 + x)5 (–x2)0 + 5C₁(1 + x)4 (–x2)1 + 5_{C2(1 + x)}3 ₍ –x2)2 + 5C₃(1 + x)2 ₍ –x2)3 + 5C₄(1 + x)1 (–x2)4 + 5C₅(1 + x)0 ₍ –x2)5  (1 + 5x + 10x2 + 10x3 + 5x4 + x5) + 5[1 + 4x + 6x2 _{+ 4x}3 _{+ x}4_]( –x2) + 10[1 + 3x + 3x2 _{+ x}3_{] (x}4₎ + 10[1 + x2 _{+ 2x] (} –x6)+ 5(1 + x) (x8) + (–x10)  (1 + 5x + 10x2 + 10x3 + 5x4 + x5) + [–5x2– 20x3– 30 x4– 20x5– 5x6] + (10x4 _{+ 30x}5 _{+ 30x}6 _{+ 10x}7₎ + [–10 x6– 10x8– 20x7] + 5x8 + 5x9– x10  –x10 + 5x9– 5x8– 10x7 + 15x6 + 11x5 – 15x4– 10x3 + 5x2 + 5x + 1 Ans. 8.  3C0(x + 2)3 º x 1        + 3_{C1(x + 2)}2 1 x 1        + 3_{C2(x + 2)}1 2 x 1        + 3C₃(x + 2)0 3 x 1         [x3 + 8 + 12x + 6x2] + 3.[x2 + 4x + 4].        x 1 + 3(x + 2). 1₂ _– 1₃

16. As T_r+1 = n r r r n_Cx _{y in (x + y)}n Now consider 17 x 7 x        [on comparing n = 17, r = 10, x = x , y = x 7  ] T₁₁ = 17C₁₀(x)17–10 10 x 7        = ₁₀ 10 7 10 17 x ) 7 ( . x C   T11 = 7 10 3 10 17 x C  _Ans. Type (III) 17. (101)100 _{= (1 + 100)}100 [(1 + x)n ₌ 0 n_{C +} 1 n_{C x +} 2 2 n_{C x +...+} n n n_{C x ]}

Using Binomial theorem (1 + 100)100 ₌ 0 100 C + 100C₁(100)1 + 100_C2(100)2 _{+... +} 100 100_{C (100)}100 Now (1+ 100)100 – 1 = 1 + 104 +100C₂104 +...10200– 1 = 104 _{[1 +} 2 100 C +... 10196_]

 1 + 100C2 +... + 10196 is a natural number by the

virtue of its being the binomial coefficients. = 104 _N  (101)100– 1 is divisible by 10,000. 18. Consider 171995 _{+ 11}1995 – 71995  (7 + 10)1995 + (1 + 10)1995– 71995 1995C₀(7)1995 + 1995C₁(7)1994(10)1 + .... 1995_C1995(10)1995 ₊ 0 1995_{C +} 1 1995_{C (10)}1 +... 1995C₁₉₉₅(10)1995 – 71995  Now 1995C₀ + 10 [ 1994 1 1995_{C (7) +...} 1994 1995 1995_{C (10)} + 1995C₁ +... 1995C₁₉₉₅(10)1994_]  1 + 10N [ 1995C1(7)1994 +... 1995 1995_{C (10)}1994 ₊ 1 1995_{C +...} 1995 1995_{C (10)}1994_]

= N(natural number as it is the sum of binomial coefficients)

 Units place is 1 Ans.

19. Consider 7 2 1 x 4 2 1         _  _{[ (x + y)}n ₌ 0 n C xn ₊ 1 n_{C x}n–1 _y1 ₊ 2 n_{C x}n–2 _y2 _{+... +} n n_{C y}n_] = 7 0 7 2 1 C       + 6 1 7 2 1 C       2 1 x 4  + 2 5 2 7 2 1 x 4 2 1 C         _       +...+ 7 7 7 2 1 x 4 C         _ ...(i) Now 7 2 1 x 4 2 1         _  = 7 0 7 2 1 C       + 6 1 7 2 1 C               _  2 1 x 4 + 2 5 2 7 2 1 x 4 2 1 C         _        +...+ 7 7 7 2 1 x 4 C         _  ...(ii) (i) – (ii) = 2[ 6 1 7 2 1 C       2 1 x 4  + 3 4 3 7 2 1 x 4 2 1 C         _       + 5 2 5 7 2 1 x 4 2 1 C         _       + 7 º 7 7 2 1 x 4 2 1 C         _       ] = 2 _4x1 [ 1 ₇ 7 2 1 . C + 3 ₇ 7 2 1 . C . (4x + 1) + 5 7 7 2 1 . C (4x + 1)2 ₊ 27 ) 1 x 4 (  3 ] = ₆ 2 1 1 x 4  [ 1 7 C + 7C₃(4x + 1) + 7C₅(4x + 1)2 _{+ (4x + 1)}3_]  1 x 4 1                    _{ }          _{ } 7 7 2 1 x 4 1 2 1 x 4 1 = 6 2 1 [7C₁ + 7C₃(4x + 1) + 7C₅(4x + 1)2 _{+ (4x + 1)}3_]  It is a polynomial of degree 3. Ans.

20. Let = x = t6 10 3 6 6 2 4 6 t t 1 t 1 t t 1 t                10 3 3 3 3 2 4 2 4 2 ) 1 t ( t ) 1 t ( ) 1 t ( 1 t t ) 1 t t ( ) 1 t (                   10 3 3 3 5 t 1 t t t         _{  }

= ! ) 1 n ( ! ) 1 n 2 ( ! ) 1 n 2 (    (–1)r –1 + _{(r 1)!(2n r)!} ! ) 1 n 2 (    (–1)2n–r = _(2n(2n_r)!(1_r)!_1)!    [(–1)r–1 + (–1)2n–r] = _(2n(2n_r)!(1_r)!_1)!                r r ) 1 ( 1 1 1 = _(2n(2n_r)!(1_r)!_1)!    [0] = 0 proved. 25. Consider 23n – 7n – 1 = (8)n– 7n – 1 = (1 + 7)n – 7n – 1 [ (1 + x)n = 0 n C + nC₁x + ... + nC_nxn] = nC₀nC₁(7)1nC₂(7)2...nC_n(7)n7n1 =17nnC₂(7)2nC₃(7)3...nC_n(7)n7n1 = nC₂(7)2nC₃(7)3...nC_n(7)n = n 2 n n 3 n 2 n 2_{[ C C 7 ... C 7} 7     ] = 49[nC₂ nC₃7 ... nC_n7n2    ] 2 n n n 3 n 2 n 7 C ... 7 . C C     = N

It is a natural number by the virtue of being a sum of binomial coefficients. 23n – 7n – 1 = 49 N  23n– 7n – 1 is divisible by 49. Proved. 26. Consider 32n+2 – 8n – 9 = (32)n+1– 8n – 9 = (9)n+1– 8n – 9 = (1 + 8)n+1 – 8n – 9 ... same = C C(8) C (8) ... Cn1(8)n1 8n 9 1 n 2 2 1 n 1 1 1 n 0 1 n             = 1 + (n + 1) 8 + n1C₂(8)2...8n18n9 = 1 + 8n + 8 + n+1_C 2 (8) 2 _{+ ...+ 8}n+1 – 8n – 9 = n+1_C 2 (8) 2 ₊ n+1_C 3 (8) 3 _{+ ... + 8}n+1 = (8)2 _[n+1_C 2 + n+1_C 3 (8) + ... + 8 n–1 _] n+1_C 2 + n+1_C 3 (8) + ... + 8 n–1 _{= N}

It is a natural number by the virtue of being a sum of binomial coefficients.  32n+2– 8n – 9 = 64N  32n+2– 8n – 9 is divisible by 64 27. L.H.S. = 5 . 4 1 – 4 . 3 1 3 . 2 1 – 2 . 1 1  + ... to  =                          5 1 – 4 1 – 4 1 – 3 1 3 1 – 2 1 – 2 1 – 1 _{+ ...} = 1 – 5 1 5 1 4 1 – 4 1 – 3 1 3 1 2 1 – 2 1     ... = 1 – 2       1–1...to 1 – 1 = 2 _         ...to 5 1 4 1 – 3 1 2 1 – 1 _{– 1}

= 2 log_e 2 – 1 = log_e4 – log_ee = log_e       e 4 = R.H.S. 28. We have ex = 1 + ! 3 x ! 2 x ! 1 x 2 3   +... to  Put x = 1, we get e = 1 + ! 3 1 ! 2 1 ! 1 1   _{+ .... } Put x = – 1, we get e–1 = 1 – _3! 1 – ! 2 1 ! 1 1  +... 

add both equation, we get e + e–1 ₌            ... ! 6 1 ! 4 1 ! 2 1 1 2 Hence ! 6 1 ! 4 1 ! 2 1 1   _{+ ....  =} 2 1 (e + e–1₎ 29. L.H.S. =             ... ! 4 x ! 3 x ! 2 x x 8 6 4 2 – _          ... ! 3 y ! 2 y y 6 4 2 = _            ... ! 3 ) x ( ! 2 ) x ( x 1 3 2 2 2 –            ... ! 3 ) y ( ! 2 ) y ( y 1 3 2 2 2 2 = ex2 –ey2 Type (IV) 30. Let log_e 10 = x Now

_

    n 0 r r r n r ) xn 1 ( x r 1 C ) 1 ( _[  log am = m log a] 



   n 0 r r r n r ) xn 1 ( 1 C ) 1 ( ₊

_

   n 0 r r r n r ) nx 1 ( rx C ) 1 ( 



   n 0 r r r n r ) nx 1 ( r C ) 1 ( + nx 1 x 



  n 1 r r r n ) 1 ( 1 r 1 n _C   1 r ) nx 1 ( r   [using nC_r = r n 1 r 1 n C_  _]

Now vi v = bd c ac b 2 2   = 2n 8 8 6 6 n 2 x x     ] C . C ) C [( ] C . C ) C [( 5 n 3 n 2 4 n 4 n 2 n 2 3 n                           ! 5 ! ) 5 n ( ! n . ! 3 )! 3 n ( ! n ! 4 ! 4 ! ) 4 n ( ! ) 4 n ( ! n ! n ! 4 ! ) 4 n ( ! n ! 2 ! ) 2 n ( ! n ! 3 ! 3 ! ) 3 n ( ! ) 3 n ( ! n ! n x 2 2  ! 4 ! 3 1 ) 3 n ( 5 1 4 ) 4 n ( 1 ! ) 5 n ( ! ) 4 n ( ! n ! n ) 2 n ( 4 1 3 ) 3 n ( 1 ! ) 4 n ( ! ) 3 n ( ! n ! n ! 3 ! 2 1 x 2 2                                              5n 15 4n 16 ) 4 n ( ) 3 n ( 5 ) 3 n ).( 2 n ( 4 . 3 9 n 3 8 n 4 ! ) 3 n ( ! ) 5 n ( ! 2 ! 4 . x 2 2                (n 1) ) 4 n ( 5 . 4 ) 2 n .( 12 ) 1 n ( ! ) 3 n ( ! ) 5 n ( ! 2 ! 4 . x 2 2 = 2 2 x  20₍(n_{n 2}4₎)_!!  

34. Let T_r , T_r+1 and T_r+2 be the three consecutive terms in the expansion of (1 + x)n As T_r+1 = nC_rxr in [1 + x]n  Tr = 1 r 1 r n x C    Tr+1 = r r n x C T_r+2 = r 1 1 r n_{C x} 

Now it is given that coefficients of

T_r , T_r+1 and T_r+2 are 76, 95, 76 repsectively.

1 r n C_ = 76 ...(i) r n_{C = 95 ...(ii)} 1 r n C_ = 76 ...(iii) Now i ii = 1 r n r n C C  = r 1 r n  = 76 95  76n – 76r + 76 = 95 r  76(n + 1) = 101 r ...(iv) ii iii = r n 1 r n C C_ = 1 r r n   = 95 76 95n – 95r = 76r + 76 95n – 76 = 101r ...(v) From (iv) 95n – 76 = 76n + 76 19n = 152 n = 8 Ans. 35. As T_r+1 = n_{Cr x}n–r _yr _{in (x + y)}n _{& consider (x + a)}n T₂ = nC₁ xn–1 _a1 _{= 240 (given) ...(i)} T₃ = nC₂ xn–2 _a2 _{= 720 ...(ii)} T₄ = n_{C3 x}n–3 _a3 _{= 1080 ...(iii)}

 

 

i ii  1 n 2 n C C . _{n 1} 2 n x x   1 2 a a = 3 = n 2 ) 1 n ( n  . x a = 3  (n – 1). x a = 6 ...(iv)

 

 

ii iii  2 n 3 n C C . _{n 2} 3 n x x   . 2 3 a a = 720 1080 = 2 3 ) 1 n ( n 6 2 ). 2 n ( ) 1 n ( n    x a = 2 3 (n – 2) x a = 2 9 ...(v)

 

 

v iv 2 n 1 n   = 9 6 .2 = 3 4 3n – 3 = 4n – 8 n = 5 From (iv) 4. x a = 6  x a = 2 3  a = 2 x 3 Put a = 2 x 3 in (i)  nC₁ xn–1 a1 = 240 5.x4_. 2 x 3 = 240 x5 _{= 32}  x = 2 Now a = 2 x 3 = 3  n = 5; a = 3; x = 2 Ans. 36. (i) Sum =

_

 1 n n t = t₁ + t₂ + t₃ + ... to  =

_

 1  n (n 1)! 1 = ! 4 1 ! 3 1 ! 2 1   _{+ ... to } =                   ... –2 ! 3 1 ! 2 1 ! 1 1 1 _{= e – 2} (ii) We have, t_n = ! ) 2 n ( 1  Sum =

_

 1  n (n 2)! 1

OBJECTIVE QUESTIONS

*

Marked Questions may have more than one correct option.

1. If the sum of the co-efficients in the expansion of (1 + 2x)n _{is 6561, then the greatest term in the expansion}

for x = 1/2 is : (1) 4th _{(2) 5}th _{(3) 6}th _{(4) none of these} 2. The expression, 6 2 2 6 2 2 1 x 2 1 x 2 2 1 x 2 1 x 2                   _{  } is a polynomial of degree (1) 5 (2) 6 (3) 7 (4) 8

3. Co-efficient of x5 _{in the expansion of (1 + x}2₎5 _{(1 + x)}4 _{is :}

(1) 40 (2) 50 (3) 30 (4) 60 4. Co-efficient of x15 _{in (1 + x +x}3 _{+ x}4₎n _{is :} (1)

_

  5 0 r r n r 3 15 n_{C C} (2)

_

 5 0 r r 5 n_C (3)

_

 5 0 r r 3 n_C (4)

_

  3 0 r r 5 n r 3 n_{C C}

5. If n is even natural and coefficient of xr _{in the expansion of}





x 1 x 1 n   is 2n_{, (|x| < 1), then} – (1) _rn/2 (2) r(n2)/2 (3) r(n2)/2 (4) rn

6. The coefficient of xn _{in polynomial (x +} 2n+1_C 0) (x + 2n+1_C 1)...(x + 2n+1_C n) is -(1) 2n + 1 (2) 22n+1 – 1 (3) 22n (4) none of these 7.

_

 n 1 r        



  1 r 0 p p p r r n_{C C 2} is equal to -(1) 4n – 3n + 1 (2) 4n– 3n– 1 (3) 4n– 3n + 2 (4) 4n– 3n 8. n_C 0– 2.3 n_C 1 + 3.3 2n_C 2– 4.3 3n_C 3 +...+ (–1) n _{(n +1)} n_C n 3 n _{is equal to} (1)





        1 2 n 3 2 1n n ₍₂₎        2 3 n 2n _{(3) 2}n _{+ 5n 2}n _{(4) (} –2)n.

9. If the sum of the coefficients in the expansion of (2 + 3cx + c2_x2₎12 _{vanishes, then c equals to}

(1) –1, 2 (2) 1, 2 (3) 1, –2 (4) –1, –2

10. The term independent of x in the expansion of ( 1 + x + 2x2₎

4 2 2 x 3 1 x 3        is (1) 10 (2) 2 (3) 0 (4) 6 11*. Let ! n 1000 a n

n  for n  N, then an is greatest, when

(1) n = 997 (2) n = 998 (3) n = 999 (4) n = 1000 12. 2k         0 n         k n –2k1 _       1 n           1 k 1 n + k 2 2          2 n           2 k 2 n –... + (– 1)k        k n          0 k n = (1) n_C k (2) n+1Ck (3) n–1Ck (4) n+2Ck