Prove that 1 2 C 1 n - 2 3 C 2 n + 3 4 C 3 n - 4 5 C 4 n + … . . + ( - 1 ) n + 1 n n + 1 · C n n = 1 n + 1

Let, 1 - x n = C 0 n - C 1 n x + C 2 n x 2 - C 3 n x 3 + . . . . . + C n n - x n Integrate both sides with respect to ' x ' 1 - x n + 1 - 1 n + 1 = C 0 n x - C 1 n x 2 2 + C 2 n x 3 3 - C 3 n x 4 4 + . . . . . + C n n - 1 n x n + 1 n + 1 ⇒ - 1 - x n + 1 x n + 1 = C 0 n - C 1 n x 2 + C 2 n x 2 3 - C 3 n x 3 4 + . . . . . + C n n - 1 n x n n + 1 Differentiate both sides with respect to ' x ' - 1 n + 1 x · n + 1 1 - x n - 1 - 1 - x n + 1 x 2 = 0 - C 1 n · 1 2 + C 2 n · 2 x 3 - C 3 n · 3 x 2 4 + . . . . . + C n n - 1 n n x n - 1 n + 1 Put x = 1 both sides, we get ⇒ 1 2 C 1 n - 2 3 C 2 n + 3 4 C 3 n - 4 5 C 4 n + . . . . . + - 1 n + 1 n + 1 = 1 n + 1

Prove that ∑ r = 0 n r 2 C r n p r q n - r = n p q + n 2 p 2 , if p + q = 1 .

Let ∑ r = 0 n r 2 C r n p r q n - r = ∑ r = 0 n r + r r - 1 C r n p r q n - r = ∑ r = 0 n r C r n p r q n - r + ∑ r = 0 n r r - 1 C r n p r q n - r = ∑ r = 1 n n p C r - 1 n - 1 p r - 1 q n - r + ∑ r = 2 n n n - 1 p 2 C r - 2 n - 2 p r - 2 q n - r = n p ∑ r - 1 n C r - 1 n - 1 p r - 1 q n - 1 - r - 1 + n n - 1 p 2 ∑ r = 2 n C r - 2 n - 2 p r - 2 q n - 2 - r - 2 = n p q + p n - 1 + n n - 1 p 2 q + p n - 2 Since it is given that p + q = 1 ⇒ ∑ r = 0 n r 2 C r n p r q n - r = n p 1 n - 1 + n 2 p 2 - n p 2 1 n - 2 = n p + n 2 p 2 - n p 2 = n p 1 - p + n 2 p 2 = n p q + n 2 p 2

Prove that : ( n - 1 ) 2 . C 1 + ( n - 3 ) 2 . C 3 + ( n - 5 ) 2 . C 5 + … . . = n ( n + 1 ) 2 n - 3

Consider the series e x + 1 n = e n x + C 1 e n - 1 x + C 2 e n - 2 x + . . . . . + C n e 0 x . . . 1 and e x - 1 n = e n x - C 1 e n - 1 x + C 2 e n - 2 x - . . . . . + C n - 1 n e 0 x . . . 2 Here C r = C r n , r = 1 , 2 , 3 , . . . . . . , n Subtracting equation 1 & 2 , we get e x + 1 n - e x - 1 n = 2 C 1 e n - 1 x + C 3 e n - 3 x + C 5 e n - 5 x + . . . . . Since we know that e x = 1 + x + x 2 2 ! + x 3 3 ! + . . . ⇒ 2 + x + x 2 2 ! + . . . n - x + x 2 2 ! + . . . n = 2 C 1 1 + n - 1 x + n - 1 2 x 2 2 ! + . . . + C 3 1 + n - 3 x + n - 3 2 2 ! x 2 + . . . + C 5 1 + n - 5 x + n - 5 2 2 ! x 2 + . . . + . . . Coefficient of x 2 in R.H.S. term = 2 C 1 · n - 1 2 + C 3 · n - 3 2 + C 5 · n - 5 2 + . . . And in L.H.S. term coefficient of x 2 = C 1 n · 2 n - 1 2 ! + C 2 n · 2 n - 2 = n · 2 n - 2 + n n - 1 2 ! × 2 n - 2 = n · 2 n - 3 n - 1 + 2 = n n + 1 2 n - 3 Therefore by comparing the coefficient of x 2 on both sides, we get n - 1 2 · C 1 + n - 3 2 · C 3 + n - 5 2 · C 5 + . . . . = n n + 1 2 n - 3

Prove that C r n + 2 . C r n + 1 + 3 . C r n + 2 + … … . + ( n + 1 ) . C r 2 n = C r + 2 n + ( n + 1 ) . 2 n + 1 C r + 1 - C r + 2 2 n + 1

Coefficient of x r in the expansion of 1 + x n is C r n Hence, given expression C r n + 2 . C r n + 1 + 3 . C r n + 2 + … … . + ( n + 1 ) . C r 2 n can be rewritten as, coefficient of x r in the expansion of 1 + x n + 2 . 1 + x n + 1 + 3 . 1 + x n + 2 + . . . . . . + n + 1 . 1 + x 2 n Let this expression be represented by E Hence, E = 1 + x n + 2 . 1 + x n + 1 + 3 . 1 + x n + 2 + . . . . . . + n + 1 . 1 + x 2 n Simplifying this by substituting 1 + x = t , we get E = t n + 2 . t n + 1 + 3 . t n + 2 + . . . . . . + n + 1 . t 2 n = t n 1 + 2 t + 3 t 2 + . . . . . + n + 1 t n Let E t n = S , Hence, S = 1 + 2 t + 3 t 2 + . . . . . + n + 1 t n ⇒ S - t S = 1 + 2 t + 3 t 2 + . . . . . + n + 1 t n - t + 2 t 2 + 3 t 3 + . . . . . + n t n + n + 1 t n + 1 ⇒ 1 - t S = 1 + t + t 2 + . . . . . + t n - n + 1 t n + 1 = t n + 1 - 1 t - 1 - n + 1 t n + 1 ∴ S = 1 - t n + 1 t - 1 2 + n + 1 t n + 1 t - 1 or, E = S . t n = t n 1 - t n + 1 t - 1 2 + n + 1 t 2 n + 1 t - 1 But t - 1 = x Hence, E = 1 + x n - 1 + x 2 n + 1 x 2 + n + 1 1 + x 2 n + 1 x Hence, required coefficient of x r is in 1 + x n x 2 + n + 1 1 + x 2 n + 1 x - 1 + x 2 n + 1 x 2 which is, C r + 2 n + ( n + 1 ) 2 n + 1 C r + 1 - C r + 2 2 n + 1

Show that 3 = 1 + 1 3 + 1 3 · 3 6 + 1 3 · 3 6 · 5 9 + 1 3 · 3 6 · 5 9 · 7 12 + … …

Given, 3 3 = 3 1 2 = 1 3 - 1 2 = 1 - 2 3 - 1 2 We know that, If x < 1 & n ∈ Q , then 1 - x n = 1 + n x + n n - 1 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + n n - 1 n - 2 n - 3 4 ! x 4 + . . . . . . . , ⇒ 1 - 2 3 - 1 2 = 1 + - 1 2 - 2 3 + - 1 2 - 1 2 - 1 2 ! - 2 3 2 + - 1 2 - 1 2 - 1 - 1 2 - 2 3 ! - 2 3 3 + - 1 2 - 1 2 - 1 - 1 2 - 2 - 1 2 - 3 4 ! - 2 3 4 + . . . . . = 1 + 1 3 + 1 2 3 4 1 2 4 9 + 1 2 3 4 5 2 1 6 8 27 + 1 2 3 4 5 2 7 2 1 24 16 81 + . . . . . . . = 1 + 1 3 + 1 3 · 3 6 + 1 3 · 3 6 · 5 9 + 1 3 · 3 6 · 5 9 · 7 12 + … …

If ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + … … + C n x n , show that for m ≥ 2 C 0 - C 1 + C 2 - … … … + - 1 m - 1 C m - 1 = - 1 m - 1 C m - 1 n - 1

Given, ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + … … + C n x n , ⇒ ( 1 - x ) n = C 0 - C 1 x + C 2 x 2 - … … + - 1 n C n x n , ( - 1 ) m - 1 C m - 1 n - 1 = C 0 - C 1 + C 2 - … … … + ( - 1 ) m - 1 C m - 1 ⇒ ( - 1 ) m - 1 C m - 1 n - 1 = Coefficient of x m - 1 in the 1 - x n - 1 = 1 - x n 1 - x - 1 ⇒ 1 - x n 1 - x - 1 = C 0 - C 1 x + C 2 x 2 - . . . . . + - 1 m - 1 C m - 1 x m - 1 + . . . . 1 + x + x 2 + . . . + x m - 1 + . . . . Coefficient of x m - 1 = C 0 - C 1 + C 2 - . . . . + - 1 m - 1 C m - 1 n - 1 .

If a 0 , a 1 , a 2 , … . . be the coefficients in the expansion of 1 + x + x 2 n in ascending powers of x , then prove that: i a 0 a 1 - a 1 a 2 + a 2 a 3 - … = 0 ii a 0 a 2 - a 1 a 3 + a 2 a 4 - … … + a 2 n - 2 a 2 n = a n + 1 iii E 1 = E 2 = E 3 = 3 n - 1 ; where E 1 = a 0 + a 3 + a 6 + … ; E 2 = a 1 + a 4 + a 7 + … & E 3 = a 2 + a 5 + a 8 + …

We have, 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + a 3 x 3 . . . . + a 2 n x 2 n . . . 1 Replace x → - 1 x , then 1 - 1 x + 1 x 2 n = a 0 - a 1 x + a 2 x 2 - a 3 x 3 + . . . + - 1 n a n x n + . . . . + a 2 n x 2 n ⇒ 1 - x + x 2 x 2 n = a 0 - a 1 x + a 2 x 2 - a 3 x 3 + . . . + - 1 n a n x n + . . . . + a 2 n x 2 n . . . 2 Multiplying i and ii , we get 1 + x + x 2 n 1 - x + x 2 n x 2 n = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . . . + a n x n + . . . . + a 2 n x 2 n a 0 - a 1 x + a 2 x 2 - a 3 x 3 + . . . + - 1 n a n x n + . . . . + a 2 n x 2 n ⇒ 1 + x 2 2 - x 2 n x 2 n = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . . . + a n x n + . . . . + a 2 n x 2 n a 0 - a 1 x + a 2 x 2 - a 3 x 3 + . . . + - 1 n a n x n + . . . . + a 2 n x 2 n . . . . . 3 i Comparing coefficient of x - 1 both sides, we get a 0 a 1 - a 1 a 2 + a 2 a 3 - … = Coefficient of x 2 n - 1 in 1 + x 4 + x 2 n . Now, replace x → x 2 , in 1 , we get 1 + x 2 + x 4 n = a 0 + a 1 x 2 + a 2 x 4 + a 3 x 6 + . . . + a n x 2 n + . . . . + a 2 n x 4 n This series has all even power terms, so coefficient of x 2 n - 1 is 0 . Hence, - a 0 a 1 + a 1 a 2 - a 2 a 3 + … = 0 ⇒ a 0 a 1 - a 1 a 2 + a 2 a 3 - … = 0 ii Comparing coefficient of x - 2 in 3 , we get a 0 a 2 - a 1 a 3 + a 2 a 4 - … … + a 2 n - 2 a 2 n = Coefficient of x - 2 in 1 + x 2 2 - x 2 n x 2 n = 1 + x 4 + x 2 n x 2 n ⇒ a 0 a 2 - a 1 a 3 + a 2 a 4 - … … + a 2 n - 2 a 2 n = Coefficient of x - 2 in a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . . . + a n x n + . . . . + a 2 n x 2 n x 2 n ⇒ a 0 a 2 - a 1 a 3 + a 2 a 4 - … … + a 2 n - 2 a 2 n = Coefficient of x 2 n - 2 in a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . . . + a n x n + . . . . + a 2 n x 2 n Replacing x → x 2 in 1 , we get a 0 a 2 - a 1 a 3 + a 2 a 4 - … … + a 2 n - 2 a 2 n = a n - 1 iii We have, E 1 = a 0 + a 3 + a 6 + … ; E 2 = a 1 + a 4 + a 7 + … & E 3 = a 2 + a 5 + a 8 + … Putting x = 1 in 1 , we get a 0 + a 1 + a 2 + a 3 + a 4 + a 5 + . . . . + a 2 n - 2 + a 2 n - 1 + a 2 n = 3 n ⇒ a 0 + a 3 + a 6 + . . . . . + a 1 + a 4 + a 7 + . . . . + a 2 + a 5 + a 8 + . . . . = 3 n ⇒ E 1 + E 2 + E 3 = 3 n . . . 4 Put x = ω (cube root of unity) in 1 , we get 1 + ω + ω 2 n = a 0 + a 1 ω + a 2 ω 2 + a 3 ω 3 . . . . + a 2 n ω 2 n ⇒ a 0 + a 1 ω + a 2 ω 2 + a 3 ω 3 + a 4 ω 4 + a 5 ω 5 + a 6 ω 6 + . . . . + a 2 n ω 2 n = 0 ⇒ a 0 + a 1 ω + a 2 ω 2 + a 3 + a 4 ω 4 + a 5 ω 5 + a 6 + . . . . + a 2 n ω 2 n = 0 ⇒ a 0 + a 3 + a 6 + . . . + ω a 1 + a 4 + a 7 + . . . . + ω 2 a 2 + a 5 + a 8 + . . . . = 0 ⇒ E 1 + ω E 2 + ω 2 E 3 = 0 ⇒ E 1 + ω E 2 + - 1 - ω E 3 = 0 ⇒ E 1 - E 3 + ω E 2 - E 3 = 0 ⇒ ω E 2 - E 3 = E 3 - E 1 Since, E 1 , E 2 , E 3 are real, and ω E 2 - E 3 , hence we must have E 1 = E 2 = E 3 Putting in 4 , we get 3 E 1 = 3 n ⇒ E 1 = 3 n - 1 ⇒ E 2 = E 3 = 3 n - 1 Hence, prove.

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4

Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 4: Exercise-4

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 4: Exercise-4

Attempt the free practice questions on Chapter 7: Binomial Theorem, Exercise 4: Exercise-4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 4: Exercise-4 with Hints & Solutions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 22

Prove that $\frac{1}{2} {}^{n}C_{1} - \frac{2}{3} {}^{n}C_{2} + \frac{3}{4} {}^{n}C_{3} - \frac{4}{5} {}^{n}C_{4} + \dots . . + \frac{(- 1)^{n + 1} n}{n + 1} \cdot {}^{n}C_{n} = \frac{1}{n + 1}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 23

Prove that $\sum_{r = 0}^{n} r^{2} {}^{n}C_{r} p^{r} q^{n - r} = n p q + n^{2} p^{2},$ if $p + q = 1 .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 24

Prove that $: (n - 1)^{2} . C_{1} + (n - 3)^{2} . C_{3} + (n - 5)^{2} . C_{5} + \dots . . = n (n + 1) 2^{n - 3}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 25

Prove that ${}^{n}C_{r} + 2 . {}^{n + 1}C_{r} + 3 . {}^{n + 2}C_{r} + \dots \dots . + (n + 1) . {}^{2 n}C_{r} = {}^{n}C_{r + 2} + (n + 1) .^{2 n + 1} C_{r + 1} - {}^{2 n + 1}C_{r + 2}$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 26

Show that $\sqrt{3} = 1 + \frac{1}{3} + \frac{1}{3} \cdot \frac{3}{6} + \frac{1}{3} \cdot \frac{3}{6} \cdot \frac{5}{9} + \frac{1}{3} \cdot \frac{3}{6} \cdot \frac{5}{9} \cdot \frac{7}{12} + \dots \dots$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 27

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots \dots + C_{n} x^{n},$ show that for $m \geq 2$
$C_{0} - C_{1} + C_{2} - \dots \dots \dots + {(- 1)}^{m - 1} C_{m - 1} = {(- 1)}^{m - 1} {}^{n - 1}C_{m - 1}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 28

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots . + C_{n} x^{n},$ then show that the sum of the products of the $C_{i}' s$ taken two
at a time, represented by $\underset{0 \leq i ⩽ j ⩽ n}{\sum \sum C_{i} C_{j}}$ is equal to $2^{2 n - 1} - \frac{(2 n)!}{2 (n!)^{2}} .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 7 - Binomial Theorem>Exercise-4>Q 29

If $a_{0}, a_{1}, a_{2}, \dots . .$ be the coefficients in the expansion of ${(1 + x + x^{2})}^{n}$ in ascending powers of $x,$ then prove that:

$(i) a_{0} a_{1} - a_{1} a_{2} + a_{2} a_{3} - \dots = 0$

$(ii) a_{0} a_{2} - a_{1} a_{3} + a_{2} a_{4} - \dots \dots + a_{2 n - 2} a_{2 n} = a_{n + 1}$

$(iii) E_{1} = E_{2} = E_{3} = 3^{n - 1};$ where $E_{1} = a_{0} + a_{3} + a_{6} + \dots; E_{2} = a_{1} + a_{4} + a_{7} + \dots & E_{3} = a_{2} + a_{5} + a_{8} + \dots$

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