Show that 3 C 0 - 8 C 1 + 13 C 2 - 18 C 3 + … + ( n + 1 ) t h term = 0 .

LHS = 3 C 0 - 8 C 1 + 13 C 2 - 18 C 3 + … + ( n + 1 ) t h . General term of the above series can be written as T r = ( - 1 ) r ( 3 + 5 r ) n C r , r = 0 , 1 , 2 , … … , n Sum of the above series can also be written as S = ∑ r = 0 n ( - 1 ) r ( 3 + 5 r ) n C r = 3 ∑ r = 0 n ( - 1 ) r n C r + 5 ∑ r = 0 n ( - 1 ) r r n C r = 3 ∑ r = 0 n ( - 1 ) r n C r + 5 ∑ r = 1 n - 1 r r n r C r - 1 n - 1 = 3 ∑ r = 0 n ( - 1 ) r n C r + 5 n ∑ r = 1 n ( - 1 ) r n - 1 C r - 1 = 3 ( 0 ) + 5 n ( 0 ) = 0 Hence, 3 C 0 - 8 C 1 + 13 C 2 - 18 C 3 + … + n + 1 t h term = 0 .

Show that C o n 2 + C 1 ( 2 - n ) 2 + C 2 ( 4 - n ) 2 + … … + C n ( 2 n - n ) 2 = n 2 n .

Given, C o n 2 + C 1 ( 2 - n ) 2 + C 2 ( 4 - n ) 2 + … … + C n ( 2 n - n ) 2 Summation of this series can be written as follows = C 0 n 2 + ∑ k = 1 n C k 4 k 2 + n 2 - 4 k n = C 0 n 2 + ∑ k = 1 n n 2 C k + 4 k 2 C k - 4 k n C k = C 0 n 2 + n 2 ∑ k = 1 n C k + 4 ∑ k = 1 n k 2 C k - 4 n ∑ k = 1 n k C k 1 2 3 1 C 0 2 + n 2 ∑ k = 1 n C k = n 2 C 0 + C 1 + C 2 + C 3 + ⋯ + C n ⇒ C 0 2 + n 2 ∑ k = 1 n C k = n 2 × 2 n · · · i To get the result for 2 ∑ k = 1 n k 2 C k = C 0 + 4 C 2 + 9 C 3 + ⋯ + n 2 C n ( 1 + x ) n = C 0 + x C 1 + C 2 x 2 + C 3 x 3 + ⋯ + C n x n Differentiating with respect to x . n ( 1 + x ) n - 1 = C 1 + 2 C 2 x + 3 C 3 x 2 + ⋯ + n C n x n - 1 Multiplying with x on both sides n x ( 1 + x ) n - 1 = C 1 x + 2 C 2 x 2 + 3 C 3 x 3 + ⋯ + n C n x n Differentiating with respect to x on both sides. n ( 1 + x ) n - 1 + n x ( 1 + x ) n - 2 ( n - 1 ) = C 1 + 2 2 C 2 x + 3 2 C 3 x 2 + · · · + n 2 C n x n - 1 Plug in x = 1 n × 2 n - 1 + n ( n - 1 ) 2 n - 2 = C 1 + 4 C 2 + 9 C 3 + ⋯ + n 2 C n C 1 + 4 C 2 + 9 C 3 + ⋯ + n 2 C n = n 2 n - 1 + n ( n - 1 ) n - 2 4 ∑ k = 1 n k 2 C k = 4 × n × 2 n - 1 + n ( n - 1 ) × 2 n - 2 × 4 4 ∑ k = 1 n k 2 C k = n × 2 n + 1 + n ( n - 1 ) 2 n · · · i i Solving 3 ∑ k = 1 n k C k = C 1 + 2 C 2 + 3 C 3 + · · · + n C n ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + ⋯ + C n x n Differentiate with respect to x . n ( 1 + x ) n - 1 = C 1 + 2 C 2 x + 3 C 3 x 2 + ⋯ + n C n x n - 1 Put x = 1 . n × 2 n - 1 = C 1 + 2 C 2 × 3 c 3 + - + n C n ∑ k = 1 n k C k = n × 2 n - 1 4 n ∑ k = 1 n k C k = 4 n n × 2 n - 1 4 n ∑ k = 1 n k C k = n 2 2 n + 1 · · · i i i Plug in solved forms of i i i & i i i in 1 2 & 3 ⇒ n 2 × 2 n + n × 2 n + 1 + n ( n - 1 ) 2 n - n 2 × 2 n + 1 ⇒ n 2 × 2 n + n × 2 n × 2 + n 2 2 n - n 2 n - n 2 2 n × 2 ⇒ n × 2 n ( 2 - 1 ) ⇒ n × 2 n Hence, C o n 2 + C 1 ( 2 - n ) 2 + C 2 ( 4 - n ) 2 + … … + C n ( 2 n - n ) 2 = n 2 n

Find C 0 + 2 C 1 + 3 C 2 + … + ( 1 ) n ( n + 1 ) C n .

We know the standard series ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + ⋯ + C n x n Multiplying with x on both sides. x ( 1 + x ) n = C 0 x + C 1 x 2 + C 2 x 3 + ⋯ + C n x n + 1 Differentiating with respect to x on both sides. ( 1 + x ) n + x n ( 1 + x ) n - 1 = C 0 + 2 C 1 x + 3 C 2 x 2 + ⋯ · + n + 1 x n C n Plug in x = 1 in the series, then we get 2 n + n ( 2 ) n - 1 = C 0 + 2 C 1 + 3 C 2 + ⋯ + ( n + 1 ) C n Hence, C 0 + 2 C 1 + 3 C 2 + ⋯ + ( n + 1 ) C n = 2 n - 1 n + 2

Show that C 0 + 3 C 1 + 5 C 2 + … . + ( 2 n + 1 ) C n = ( n + 1 ) 2 n .

Consider the following series 1 + x 2 n = 1 + C 1 n x 2 + C 2 n x 4 + C 3 n x 6 + … + C n n x 2 n Multiplying with x on both sides ⇒ x 1 + x 2 n = x + C 1 n x 3 + C 2 n x 5 + C 3 n x 7 + … + C n n x 2 n + 1 Now differentiating both sides with respect to x and plug in x = 1 . 1 + x 2 n + n x ( 2 x ) 1 + x 2 n - 1 x = 1 = 1 + 3 n C 1 + 5 n C 2 + … + 2 n + 1 n C n ⇒ 1 + 3 n C 1 + 5 n C 2 + … + ( 2 n + 1 ) C n n = 1 + x 2 n + n x ( 2 x ) 1 + x 2 n - 1 x = 1 = 2 n + 2 n 2 n - 1 = 2 n + n 2 n = 2 n ( n + 1 ) Hence, 1 + 3 n C 1 + 5 n C 2 + … ( 2 n + 1 ) C n n = 2 n ( n + 1 ) .

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)

Odisha Board Solutions for Chapter: Binomial Theorem, Exercise 2: EXERCISES - 9 (b)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Binomial Theorem, Exercise 2: EXERCISES - 9 (b)

Attempt the free practice questions on Chapter 9: Binomial Theorem, Exercise 2: EXERCISES - 9 (b) with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.