60 Binomial Theorem Part 1 of 2

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C O M P L E X N U M B E R S / P a g e 1 o f 2 5 T E K O C L A S S E S , H .O .D . M A T H S : S U H A G R . K A R IY A ( S . R . K . S ir ) P H : (0 7 5 5 3 2 0 0 0 0 0 , 0 9 8 9 3 0 5 8 8 8 1 , B H O P A L , ( M .P .)

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Index

1. Theory

2. Short Revision

3. Exercise (Ex. 1 to 8)

4. Assertion & Reason

5. Que. from Compt. Exams

6. 34 Yrs. Que. from IIT-JEE

7. 10 Yrs. Que. from AIEEE

Subject : Mathematics

Topic :

Binomial Theorem

Student’s Name :______________________

Class

:______________________

Roll No.

:______________________

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Binomial Theorem

1 .

Bino mial Expression :

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 3 1/3 ) 1 x ( 1 + etc.

2 .

Statement of Binomial theorem :

If a, b ∈ R and n ∈ N, then ; (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 or (a + b)n₌

∑

= − n 0 r r r n r n b a C

Now, putting a = 1 and b = x in the binomial theorem -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 (1 + x)n₌

∑

= n 0 r r r n x C

Solved Example # 1: Expand the following binomials :

(i) (x – 3)5 _(ii) 4 2 2 x 3 1 _       −−−− Solution. (i) (x – 3)5₌5_C 0x 5₊5_C 1x 4_{(– 3)}1₊5_C 2 x 3_{(– 3)}2₊5_C 3 x 2_(–3)3 + 5_C 4 x (– 3) 4₊5_C 5 (– 3) 5 = x5_{– 15x}4_{+ 90x}3_{– 270x}2_{+ 405x – 243} (ii) 4 2 2 x 3 1 _       − ₌4_C 0 + 4_C 1         − 2 x 3 2 + 4_C 2 2 2 2 x 3         − + 4_C 3 3 2 2 x 3         − ₊4_C 4 4 2 2 x 3         − = 1 – 6x2₊ 2 27 x4_– 2 27 x6₊ 16 81 x8

Solved Example # 2: Expand the binomial

20 2 y 3 3 x 2 _      ++++ up to four terms Solution. 20 2 y 3 3 x 2 _      ₊ = 20_C 0 20 3 x 2 _      + 20_C 1 19 3 x 2 _            2 y 3 + 20_C 2 18 3 x 2 _      2 2 y 3 _      + 20_C 3 17 3 x 2 _      3 2 y 3 _      + .... = 20 3 x 2 _      + 20. 18 3 2_      x19_{y + 190 .} 16 3 2_      x18_y2_{+ 1140} 14 3 2_      x17_y3_{+ ...}

Self practice problems

1. Write the first three terms in the expansion of

6 3 y 2       − _.

2. Expand the binomial

5 2 x 3 3 x         + _. Ans. (1) 64 – 64y + 3 80 y2 ₍₂₎ 243 x10 + 27 5 x7₊ 3 10 x4_{+ 30x +} 2 x 135 + ₅ x 243 .

3 .

Properties of Binomial Theorem :

(i) The number of terms in the expansion is n + 1. (ii) The sum of the indices of x and y in each term is n. (iii) The binomial coefficients (n_C

0,

n_C

1 ...

n_C

n) of the terms equidistant from the beginning and

the end are equal, i.e. n_C

0 = n_C n, n_C 1 = n_C n–1 etc. {∵ n_C r = n_C n–r}

Solved Example # 3: The number of dissimilar terms in the expansion of (1 – 3x + 3x2_{– x}3₎20_is

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

Solution. (1 – 3x + 3x2_{– x}3₎20_{= [(1 – x)}3_]20_{= (1 – x)}60

Therefore number of dissimilar terms in the expansion of (1 – 3x + 3x2_{– x}3₎20_is_61.

4 .

Some important terms in the expansion of (x + y)

nnnn

_:

(i) 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. T_r+1 = n_C

r x

n–r_yr

Solved Example # 4

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

9 x 2 5 5 x 4 _      −−−−

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Solution. (i) T_{27 + 1} = 30_C 27 (5x) 30– 27_(8y)27 = ! 27 ! 3 ! 30 (5x)3_{. (8y)}27 _Ans. (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 _      − = ₃9_!₆! _! 3 5 x 4 _      6 x 2 5 _      = 3 x 10500 Ans.

Solved 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

(

₉1/4₊₈1/6

)

1000_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 integres 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 Ans.

(ii) Middle term (s) :

(a) If n is even, there is only one middle term, which is _n+₂2_th term.

(b) If n is odd, there are two middle terms, which are _n₂+1_th and _n₂+1+1_th terms.

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

(i) 14 2 2 x 1 _       −−−− _(ii) 9 3 6 a a 3 _       −−−− Solution. (i) 14 2 2 x 1 _       −

Here, n is even, therefore middle term is       + 2 2 14 th term. It means T₈ is middle term

T₈ = 14_C 7 7 2 2 x         − = – 16 1616 16 429 429 429 429 x14_{. Ans.} (ii) 9 3 6 a a 3 _       −

Here, n is odd therefore, middle terms are _9₂+1_th & _9₂+1+1_th. It means T₅ & T₆ are middle terms

T₅ = 9_C 4 (3a) 9 – 4 4 3 6 a         − ₌ 8888 189 189 189 189_a17 _Ans. T₆ = 9_C 5 (3a) 9 – 5 5 3 6 a         − ₌_– 16 16 16 16 21 2121 21 a19_. _Ans.

(iii) Term containing specified powers of x in

n x b ax       _± β α

Solved Example # 7: Find the coefficient of x32_{and x}–17_in

15 3 4 x 1 x       −−−− .

Solution.: Let (r + 1)th term contains xm

T_{r + 1} = 15_C r (x 4₎15 – r r 3 x 1      ₋ = 15_C r x 60 – 7r_{(– 1)}r (i) for x32_{, 60 – 7r = 32} ⇒ 7r = 28 ⇒ r = 4. (T₅) T₅ = 15_C 4 x 32_{(– 1)}4

Hence, coefficient of x32_is_1365Ans.

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⇒ r = 11 (T₁₂) T₁₂= 15_C 11 x –17_{(– 1)}11

Hence, coefficient of x–17_is_{– 1365} _Ans.

(iv) Numerically greatest term in the expansion of (x + y)n_{, n}∈∈ N∈∈

Let T_r and T_r+1 be the rth and (r + 1)th terms respectively T_r = n_C r–1 x n–(r–1)_yr–1 T_r+1 = n_C r x n–r_yr Now, r 1 r T T₊ = n r 1 r 1 r r n 1 r n r n y x y x C C − + − − − = r 1 r n− + . _xy Consider r 1 r T T₊ ≥ 1       − + r 1 r n x y ≥ 1 r 1 n+ – 1 ≥ y x r ≤ y x 1 1 n + + Case - ΙΙΙΙ When y x 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 y x 1 1 n + +

is an integer, equal to m, then TT_m and T_m+1 will be numerically greatest terms

(both terms are equal in magnitude)

Case - ΙΙΙΙΙΙΙΙ When y x 1 1 n + +

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

(i) T_r+1 > T_r when r < y x 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 > y x 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 y x 1 1 n + +

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

greatest term.

Solved 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 ≥ 15_C r – 1 3 15 – (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 Explanation: For r ≤ 4, T_{r + 1} ≥ T_r ⇒ T₂ > T₁ T₃ > T₂ T₄ > T₃ T₅ = T₄ For r > 5, T_{r + 1} < T_r T₆ < T₅ T₇ < T₆

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and so on Hence, T₄ and T₅ are numerically greatest terms and both are equal.

Self practice problems :

3. Find the term independent of x in

9 2 x 3 x       ₋

4. The sum of all rational terms in the expansion of (31/5_{+ 2}1/3₎15_is

(A) 60 (B) 59 (C) 95 (D) 105

5. Find the coefficient of x–1_{in (1 + 3x}2_{+ x}4₎

8 x 1 1       +

6. Find the middle term(s) in the expansion of (1 + 3x + 3x2_{+ x}3₎2n

7. Find the numerically greatest term in the expansion of (7 – 5x)11_{where x =}

3 2 . Ans. (3) 28.37 ₍₄₎ _B ₍₅₎ ₂₃₂ (6) 6n_C 3n . x 3n ₍₇₎ _T 4 = ₉ 440 × 78_{× 5}3_.

5 .

Multino mial Theo rem

: As we know the Binomial Theorem –

(x + y)n₌

∑

∑

= + +r r 11 r1 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 + ₉(11_!₂)!_! . ₁2_!₁!_! 71₊ ! 4 ! 7 )! 11 ( . ₂4_!₂!_! 72₊ ! 6 ! 5 ! ) 11 ( . ₃6_!₃!_! 73 + ₃(11_!₈)!_! . ₄8_!₄! _! 74₊ ! 10 ! 1 ! ) 11 ( . ₅(10_!₅)_!! 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 _Ans.

8. The number of terms in the expansion of (a + b + c + d + e + f)n_is

(A) n+4_C 4 (B) n+3_C n (C) n+5_C n (D) n + 1

9. Find the coefficient of x3_y4_z2_{in the expansion of (2x – 3y + 4z)}9

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Ans. (8) C (9) ₃_!9₄!_!₂_! (10) 23₃4₄2_{– 91}

6 .

Application of Binomial Theorem :

(i) If ( A+B)n = ΙΙΙΙ + f where ΙΙΙΙ and n are positive integers, n being odd and 0 < f < 1 then (ΙΙΙΙ + f) f = kn_{where A – B}2_{= k > 0 and} _A _{– B < 1.}

If n is an even integer, then (ΙΙΙΙ + f) (1 – f) = kn

If n is positive integer, then prove that the integral part of (7 + 4 3 )n_{is an odd number..}

Solution. Let (7 + 4 3 )n = Ι + f _...(i)

where Ι & f are its integral and fractional parts respectively. It means 0 < f < 1

Now, 0 < 7 – 4 3 < 1 0 < (7 – 4 3 )n_{< 1}

Let (7 – 4 3 )n = f′ _...(ii)

⇒ 0 < f′ < 1 Adding (i) and (ii)

Ι + f + f′ = (7 + 4 3 )n_{+ (7 – 4 3 )}n = 2 [n_C 0 7 n₊n_C 2 7 n – 2_{(4 3 )}2_{+ ...]}

Ι + f + f′ = even integer(f + f′ must be an integer) 0 < f + f′ < 2 ⇒ f + f′ = 1

Ι + 1 = even integer therefore Ι is an odd integer.

Show that the integer just above ( 3 + 1)2n_{is divisible by 2}n + 1_{for all n}∈∈ N.∈∈

Solution. Let ( 3 + 1)2n_{= (4 + 2 3 )}n_{= 2}n_{(2 + 3 )}n = Ι + f _...(i)

where Ι and f are its integral & fractional parts respectively 0 < f < 1.

Now 0 < 3 – 1 < 1 0 < ( 3 – 1)2n_{< 1}

Let ( 3 – 1)2n_{= (4 – 2 3 )}n_{= 2}n_{(2 – 3 )}n = f′. _...(ii)

0 < f′ < 1 adding (i) and (ii)

Ι + f + f′ = ( 3 + 1)2n_{+ ( 3 – 1)}2n = 2n_{[(2 + 3 )}n_{+ (2 – 3 )}n_] = 2.2n_[n_C 0 2 n₊n_C 2 2 n – 2_{( 3 )}2_{+ ...]}

Ι + f + f′ =2n + 1_{k (where k is a positive integer)}

0 < f + f′ < 2 ⇒ f + f′ = 1 Ι + 1 = 2n + 1_k.

Ι + 1 is the integer just above ( 3 + 1)2n_{and which is divisible by 2}n + 1_.

(ii) Cheking divisibility

Solved Example # 13: Show that 9n_{+ 7 is divisible by 8, where n is a positive integer.}

Solution. 9n_{+ 7 = (1 + 8)}n_{+ 7} = n_C 0 + n_C 1 . 8 + n_C 2 . 8 2_{+ ... +}n_C n 8 n_{+ 7.} = 8. C₁ + 82_{. C} 2 + ... + Cn . 8 n_{+ 8.}

= 8λ where, λ is a positive integer, Hence, 9n_{+ 7 is divisible by 8.}

(iii) Finding remainder Solved Example # 14

What is the remainder when 599_{is divided by 13.}

Solution.: 599 _{= 5.5}98_{= 5. (25)}49 = 5 (26 – 1)49 = 5 [49_C 0 (26) 49_–49_C 1 (26) 48_{+ ... +}49_C 48 (26) 1_–49_C 49 (26) 0_] = 5 [49_C 0 (26) 49_–49_C 1 (26) 48_{+ ...+}49_C 48 (26) 1_{– 1]} = 5 [49_C 0 (26) 49_–49_C 1(26) 48_{+ ... +}49_C 48 (26) 1_{– 13] + 60}

= 13 (k) + 52 + 8 (where k is a positive integer) = 13 (k + 4) + 8 Hence, remainder is 8. Ans.

(iv) Finding last digit, last two digits and last there digits of the given number. Solved Example # 15: Find the last two digits of the number (17)10.

Solution. (17)10 _{= (289)}5 = (290 – 1)5 = 5_C 0 (290) 5_–5_C 1 (290) 4_{+ ... +}5_C 4 (290) 1_–5_C 5 (290) 0 = 5_C 0 (290) 5_–5_C 1 . (290) 4_{+ ...}5_C 3 (290) 2_{+ 5 × 290 – 1}

= A multiple of 1000 + 1449 Hence, last two digits are 49 Ans. Note : We can also conclude that last three digits are 449.

(v) Comparison between two numbers

Solved Example # 16 : Which number is larger (1.01)1000000_{or 10,000 ?}

Solution.: By Binomial Theorem (1.01)1000000 _{= (1 + 0.01)}1000000

= 1 + 1000000_C

1 (0.01) + other positive terms

= 1 + 1000000 × 0.01 + other positive terms

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11. If n is positive integer, prove that the integral part of (5 5 + 11)2n + 1_{is an even number..}

12. If (7 + 4 3 )n = α + β, where α is a positive integer and β is a proper fraction then prove that

(1 – β) (α + β) = 1.

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

14. What is the remainder when 7103_{is divided by 25 .}

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

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

Ans. (14) 18 (15) 1, 01, 001 (16) (1.2)4000_.

7.

7. Pro perties of Binomial Coefficients :

Pro perties of Binomial Coefficients :

(1 + x)n_{= C} 0 + C1x + C2x 2_{+ ... + C} r x r_{+ ... + C} nx n _...(1)

(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_C ₂

(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 + ... = 2 n–1 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_)! ) 1 r n ( r ) 1 n ( + − + = ₍_n(₋n_r+₊1₁)!_)!_r_! = n+1_C r = R.H.S.

(5) Ratio of two consecutive binomial coefficients

1 r n r n C C − = r 1 r n− + (6) 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 − − − − − − Solved Example # 17 If (1 + x)n_{= C} 0 + C1x + C2x 2_{+ ... + c} nx

n_{, then show that}

(i) C₀ + 3C₁ + 32_C 2 + ... + 3 n_C n = 4 n_. (ii) C₀ + 2C₁ + 3. C₂ + ... + (n + 1) C_n = 2n – 1_{(n + 2).} (iii) C₀ – 2 C1 + 3 C2 – 4 C3 + ... + ( –1)n 1 n Cn ++++ = n 1 1 ++++ . Solution. (i) (1 + x)n_{= C} 0 + C1 x + C2x 2_{+ ... + C} nx n put x = 3 C₀ + 3 . C₁ + 32_{. C} 2 + ... + 3 n_{. C} n= 4 n

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=

∑

= − n 0 r r ) 1 ( _. 1 r C_r n + = 1 n 1 +

∑

= − n 0 r r ) 1 ( _. n + 1_C r + 1 _     = + + + + 1 r 1 n r n C C . 1 r 1 n = 1 n 1 + [n + 1C1 – n + 1_C 2 + n + 1_C 3 – ...+ (– 1) n_.n + 1_C n + 1] = 1 n 1 + [– n + 1C0 + n + 1_C 1 – n + 1_C 2 + ... + (– 1) n_.n + 1_C n + 1 + n + 1_C 0] = 1 n 1 + = R.H.S.

{

−n+1C0+n+1C1−n+1C2+...+(−1)n n+1Cn+1=0

}

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

Integrating both sides, with in 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 C_n + = n 1 1 + Proved Solved Example # 18 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.

= − − 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. ΙΙ ΙΙ ΙΙ ΙΙ 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 Solved Example # 19

Find the summation of the following series – (i) m_C m + m+1_C m + m+2_C m + ... + n_C m (ii) n_C 3 + 2 . n+1_C 3 + 3. n+2_C 3 + ... + n . 2n–1_C 3

Solution. (i) Ι Ι Ι Ι Method : Using property, n_C r + n_C r–1 = n+1_C r m_C m + m+1_C m + m+2_C m + ... + n_C m = m 1 m 1 m 1 m _C + _C + + ₊ + m+2_C m + ... + n_C m {∵ m_C m = m+1_C m+1}

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= m 2 m 1 m 2 m _C + _C + + ₊ + ... + n_C m = m+3_C m+1 + ... + n_C m = n_C m+1 + n_C m = n+1_C m+1 Ans. ΙΙ ΙΙ ΙΙ ΙΙ Method m_C m + m+1_C m + m+2_C m + ... + n_C m

The above series can be obtained by writing the coefficient of xm_in

(1 + x)m_{+ (1 + x)}m+1_{+ ... + (1 + x)}n Let S = (1 + x)m_{+ (1 + x)}m+1_{+... + (1 + x)}n =

(

)

[

]

x 1 x 1 ) x 1 ( + m + n−m+1− =

(

)

(

)

x x 1 x 1+ n+1− + m xm_{: S (coefficient of x}m_{in S)} xm_: x ) x 1 ( ) x 1 ( + n+1− + m

Hence, required summation of the series is n+1_C

m+1 Ans. (ii) n_C 3 + 2 . n+1_C 3 + 3 . n+2_C 3 + ... + n . 2n–1_C 3

The above series can be obatined by writing the coefficient of x3_in

(1 + x)n_{+ 2 . (1 + x)}n+1_{+ 3 . (1 + x)}n+2_{+ ... + n . (1 + x)}2n–1

Let S = (1 + x)n_{+ 2 . (1 + x)}n+1_{+ 3. (1 + x)}n+2_{+ ... + n (1 + x)}2n–1 _...(i)

(1 + x)S = (1 + x)n+1_{+ 2 (1 + x)}n+2_{+ ... + (n – 1) (1 + x)}2n–1

+ n(1 + x)2n _...(ii)

Subtracting (ii) from (i)

– xS = (1 + x)n_{+ (1 + x)}n+1_{+ (1 + x)}n+2_{+ ... + (1 + x)}2n–1_{– n(1 + x)}2n =

[

]

x 1 ) x 1 ( ) x 1 ( + n + n− – n (1 + x)2n S = ₂ n n 2 x ) x 1 ( ) x 1 ( + + + − + x ) x 1 ( n + 2n x3_{: S} _{(coefficient of x}3_{in S)} x3_: 2 n n 2 x ) x 1 ( ) x 1 ( + + + − + x ) x 1 ( n + 2n

Hence, required summation of the series is – 2n_C 5 +

n_C 5 + n .

2n_C

4 Ans.

Self practice problems : 17. Prove the following

(i) C₀ + 3C₁ + 5C₂ + ... + (2n + 1) C_n = 2n_{(n + 1)} (ii) 4C₀ + 2 42 . C₁ + 3 43 C₂ + ... + 1 n 4n 1 + + C_n = 1 n 1 5n 1 + − + (iii) n_C 0 . n+1_C n + n_C 1 . n_C n–1 + n_C 2 . n–1_C n–2 + ... + n_C n . 1_C 0 = 2 n–1_{(n + 2)} ( i v ) ( i v ) ( i v ) ( i v ) 2C 2 + 3_C 2 + ... + n_C 2 = n+1_C 3

8. Binomial Theorem For Negative Integer Or Fractional Indices

If n ∈ R then, (1 + x)n_{= 1 + nx +} ! 2 ) 1 n ( n − x2₊ ! 3 ) 2 n )( 1 n ( n − − x3_{+ ...} ... + n(n−1)(n−2_r)...(_! n−r+1) xr_{+ ...}∞.

Remarks:(i) The above expansion is valid for any rational number other than a whole number if | x | < 1. (ii) When the index is a negative integer or a fraction then number of terms in the expansion of

(1 + x)n_{is infinite, and the symbol}n_C

r cannot be used to denote the coefficient of the general term.

(iii) The first terms must be unity in the expansion, when index ‘n’ is a negative integer or fraction

(x + y)n₌          <         +       − + + =       + <         +       − + + =       + 1 y x if ... y x ! 2 ) 1 n ( n y x . n 1 y y x 1 y 1 x y if ... x y ! 2 ) 1 n ( n x y . n 1 x x y 1 x 2 n n n 2 n n n

(iv) The general term in the expansion of (1 + x)n_{is T}

r+1 = _r_! ) 1 r n ( )... 2 n )( 1 n ( n − − − + xr

(v) When ‘n’ is any rational number other than whole number then approximate value of (1 + x)n_is

1 + nx (x2_{and higher powers of x can be neglected)}

(vi) Expansions to be remembered (|x| < 1)

(a) (1 + x)–1_{= 1 – x + x}2_{– x}3_{+ ... + (–1)}r_xr + ...∞

(b) (1 – x)–1_{= 1 + x + x}2_{+ x}5_{+ ... + x}r + ...∞

(c) (1 + x)–2_{= 1 – 2x + 3x}2_{– 4x}3_{+ ... + (–1)}r_{(r + 1) x}r + ...∞

(d) (1 – x)–2_{= 1 + 2x + 3x}2_{+ 4x}3_{+ ... + (r + 1)x}r + ... ∞

Solved Example # 20: Prove that the coefficient of xr_{in (1 – x)}–n_isn+r–1_C r

Soltion.: (r + 1)th term in the expansion of (1 – x)–n_{can be written as}

T_{r +1} = ! r ) 1 r n )...( 2 n )( 1 n ( n− − − − − − + − (–x)r

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C O M P L E X N U M B E R S / P a g e 1 0 o f 2 5 T E K O C L A S S E S , H .O .D . M A T H S : S U H A G R . K A R IY A ( S . R . K . S ir ) P H : (0 7 5 5 3 2 0 0 0 0 0 , 0 9 8 9 3 0 5 8 8 8 1 , B H O P A L , ( M .P .)

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= (–1)r ! r ) 1 r n )...( 2 n )( 1 n ( n + + + − (–x)r = ! r ) 1 r n )...( 2 n )( 1 n ( n + + + − xr = ! r ! ) 1 n ( ) 1 r n )...( 1 n ( n )! 1 n ( − − + + − xr Hence, coefficient of xr_is ! r )! 1 n ( )! 1 r n ( − − + = n+r–1_C r Proved

Solved Example # 21: If x is so small such that its square and higher powers may be neglected then

find the value of 1/2

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} 24 24 24 41 4141 41 x Ans. Self practice problems :

18. Find the possible set of values of x for which expansion of (3 – 2x)1/2_{is valid in ascending powers of x.}

19. If y = 5 3 + ! 2 3 . 1 2 5 2_      + ! 3 5 . 3 . 1 3 5 2_     

+ ..., then find the value of y2_{+ 2y}

20. The coefficient of x100_in 2 ) x 1 ( x 5 3 − − is (A) 100 (B) –57 (C) –197 (D) 53 Ans. (18) x ∈ _     − 2 3 , 2 3 (19) 4 (20) C

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Short Revision

BINOMIAL EXPONENTIAL & LOGARITHMIC SERIES

1. BINOMIAL THEOREM : The formula by which any positive integral power of a binomial

expression can be expanded in the form of a series is known as B

INOMIAL

T

HEOREM

.

If x

, y ∈ R and n ∈ N, then ;

=

∑

= n 0 r n

_C

r

x

n – r

y

r

.

This theorem can be proved by Induction .

OBSERVATIONS :

(i)

The number of terms in the expansion is (n + 1) i.e. one or more than the index .

(ii)

The sum of the indices of x & y in each term is n .

(iii)

The binomial coefficients of the terms

n

_C

0

,

n

C

1

.... equidistant from the beginning and the end are

equal.

2. IMPORTANT TERMS IN THE BINOMIAL EXPANSION ARE :

(i)

General term

(ii)

Middle term

(iii)

Term independent of x &

(iv)

Numerically greatest term

(i)

The general term or the (r + 1)

th

_{term in the expansion of (x + y)}

n

_{is given by ;}

T

_r+1

=

n

_C

r

x

n−r

. y

r

(ii)

The middle term(s) is the expansion of (x + y)

n

_{is (are) :}

(iii)

Term independent of x contains no x ; Hence find the value of r for which the exponent of x is zero.

(iv)

To find the Numerically greatest term is the expansion of (1 + x)

n

_{, n}

∈ N find

x

r

1 r

n

x

C

x

C

T

1 r 1 r n r r n r 1 r

=

−

+

− −

+

_{. Put the absolute value of x & find the value of r Consistent with the}

inequality

r 1 r

T

₊

> 1.

Note that the Numerically greatest term in the expansion of (1 − x)

n

, x > 0 , n ∈ N is the same

as the greatest term in (1 + x)

n

_.

3. If

(

_A

₊

_B

)

n

= I + f, where I & n are

positive

integers, n being odd and 0 < f < 1, then

(I + f) . f = K

n

where A − B

2

_{= K > 0 & A}

− B < 1.

If n is an even integer, then (I + f) (1 − f) = K

n

_.

n

=

( ) ! ! ! 2 n n n

(iv)

C

)!

r

n

(

)

r

n

(

)!

n

2 (

−

+

REMEMBER :

(i)

(2n)! = 2

n

. n! [1. 3. 5 ... (2n − 1)]

5. BINOMIAL THEOREM FOR NEGATIVE OR FRACTIONAL INDICES :

If n ∈ Q , then (1 + x)

n

₌

+

−

+

−

_x

+

_...

∞

!

3 )

2 n

(

)

1 n

(

n

x

!

2 )

1 n

(

n

x

n

1

2 3

Provided | x | < 1.

− x

3

_{+ x}

4

− .... ∞

_{(b) (1}

− x)

−1

_{= 1 + x + x}

2

_{+ x}

3

_{+ x}

4

+ .... ∞

(c) (1 + x)

−2

= 1 − 2x + 3x

2

− 4x

3

+ .... ∞

_{(d) (1}

− x)

−2

_{= 1 + 2x + 3x}

2

_{+ 4x}

3

+ ... ∞

(iv)

The expansions in ascending powers of x are only valid if x is ‘small’. If x is large i.e. | x | > 1 then

we may find it convinient to expand in powers of

x

1 , which then will be small.

6. APPROXIMATIONS :

(1 + x)

n

_{= 1 + nx +}

2 .

1 )

1 n

(

n

−

x² +

3 .

2 .

1 )

2 n

(

)

1 n

(

n

−

x

3

_...

If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be

reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be

so small that its squares and higher powers may be neglected then (1 + x)

n

_{= 1 + nx, approximately.}

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This is an approximate value of (1 + x)

n

_.

7. EXPONENTIAL SERIES :

(i)

e

x

_{= 1 +}

+

_...

∞

!

3 x

!

2 x

!

1 x

2 3

; where

x

may be any real or complex & e =

Limit_n_{→ ∞} 1+ 1   _ n n

(ii)

a

x

_{= 1 +}

+

_n

_a

+

_...

∞

!

3 x

a

n

!

2 x

a

n

!

1 x

2 ₂ 3 ₃

l

where a > 0

Note :

(a)

e = 1 +

+

...

∞

!

3

1 !

2

1 !

1

1 (b)

e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is

2.7182818284.

(c)

e + e

−1

= 2













∞

+

...

!

6

1 !

4

1 !

2

1

1 (d)

e − e

−1

= 2













∞

+

...

!

7

1 !

5

1 !

3

1

1 (e)

Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier,

their inventor.

They are also called Natural Logarithm.

8. LOGARITHMIC SERIES :

(i)

l

n (1+ x) = x −

+

−

+

...

∞

4 x

3 x

2 x

2 3 4

_{where −1 < x ≤ 1}

(ii)

l

n (1− x) = − x −

−

+

...

∞

4 x

3 x

2 x

2 3 4

where −1 ≤ x < 1

(iii)

l

n

)

x

1 (

)

x

1 (

−

+

= 2

_













∞

+

...

5 x

3 x

x

5 3

x < 1

R

EMEMBER

:

(a)

1 −

4

1

3

1

2

1 ₊

₋

+... ∞ = ln 2

(b)

e

ln x

_{= x}

(c)

l

n2 = 0.693

(d)

l

n10 = 2.303

EXERCISE - 1

Q.1 Find the coefficients : (i) x

7

_in

11 2

x

b

1 x

a

_











+

(ii) x

−7

in

11 2

x

b

1 ax

_













−

(iii) Find the relation between a & b , so that these coefficients are equal.

Q.2 If the coefficients of (2r + 4)

th

, (r − 2)

th

_{terms in the expansion of (1 + x)}

18

_{are equal , find r.}

x

2

3

3 x













+

(b)

8 5 / 1 3 / 1

_x

x

2

1 







+

−

Q.5 Find the sum of the series

∑

=

−

n 0 r n r r

C

.

)

1 (













+

...

up

to

m

terms

2

15

2

7

n

, n ∈ N,

prove that

_a

₊

a

_b

+

_c

₊

c

_d

=

_b

2 ₊

b

_c

.

Q.9 Find the value of x for which the fourth term in the expansion,

8 log 3 x1 5 5 7 2 44 4 log

5

1

5

x 5 2













+

+ + −

n

C

r+1

.

Q.11 (a) Which is larger : (99

50

_{+ 100}

50

_{) or (101)}

50

_.

>

_n

₁

,

n

N

,

n

2 n

4 _∈

_>

+

Q.12 In the expansion of

1 7 11 + +       x

x

find the term not containing x.

Q.13 Show that coefficient of x

5

_{in the expansion of (1 + x²)}

5

_{. (1 + x)}

4

_{is 60.}

Q.14 Find the coefficient of x

4