C 0 1 + C 2 3 + C 4 5 + C 6 7 +..........= Options: A . &nbsp 2 n + 1 n + 1 B . &nbsp 2 n + 1 − 1 n + 1 C . &nbsp 2 n n + 1 D . &nbsp None of these

Answer: Option C : C Putting the values of C 0 , C 2 , C 4 ........., we get = 1 + n ( n − 1 ) 3.2 ! + n ( n − 1 ) ( n − 2 ) ( n − 3 ) 5.4 ! + ......... = 1 n + 1 [(n+1) + ( n + 1 ) n ( n − 1 ) 3 ! + ( n + 1 ) n ( n − 1 ) ( n − 2 ) ( n − 3 ) 5 ! + ..........] Put n+1 = N = 1 N [N + N ( N − 1 ) ( N − 2 ) 3 ! + N ( N − 1 ) ( N − 2 ) ( N − 3 ) ( N − 4 ) 5 ! + .........] = 1 N { N C 1 + N C 3 + N C 5 +..........} = 1 N { 2 N − 1 } = 2 n n + 1 { ∵ N = n+1} Trick: Put n = 1, then S 1 = 1 C 0 1 = 1 1 = 1 At n = 2, S 2 = 2 C 0 1 + 2 C 2 3 = 1 + 1 3 = 4 3 Also (c) ⇒ S 1 = 1, S 2 = 4 3

C01 + C23 + C45 + C67 +..........=

Discussion Forum : Binomial Theorem

Question -

$\frac{C_{0}}{1}$ + $\frac{C_{2}}{3}$ + $\frac{C_{4}}{5}$ + $\frac{C_{6}}{7}$ +..........=

Options:

A .

\frac{2^{n + 1}}{n + 1}

B .

\frac{2^{n + 1} - 1}{n + 1}

C .

\frac{2^{n}}{n + 1}

D . None of these

Answer: Option C
:
C

Putting the values of $C_{0}, C_{2}, C_{4}$ ........., we get

= 1 + $\frac{n (n - 1)}{3.2!}$ + $\frac{n (n - 1) (n - 2) (n - 3)}{5.4!}$ + .........

= $\frac{1}{n + 1}$ [(n+1) + $\frac{(n + 1) n (n - 1)}{3!}$ + $\frac{(n + 1) n (n - 1) (n - 2) (n - 3)}{5!}$ + ..........]

Put n+1 = N

= $\frac{1}{N}$ [N + $\frac{N (N - 1) (N - 2)}{3!}$ + $\frac{N (N - 1) (N - 2) (N - 3) (N - 4)}{5!}$ + .........]

= $\frac{1}{N}$ { $^{N} C_{1}$ + $^{N} C_{3}$ + $^{N} C_{5}$ +..........}

= $\frac{1}{N}$ { $2^{N - 1}$ } = $\frac{2^{n}}{n + 1}$ { ∵ N = n+1}

Trick: Put n = 1, then $S_{1}$ = $\frac{^{1} C_{0}}{1}$ = $\frac{1}{1}$ = 1

At n = 2, $S_{2}$ = $\frac{^{2} C_{0}}{1}$ + $\frac{^{2} C_{2}}{3}$ = 1 + $\frac{1}{3}$ = $\frac{4}{3}$

Also (c) ⇒ $S_{1}$ = 1, $S_{2}$ = $\frac{4}{3}$

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