Prove that `(.^(n)C_(0))/(1)+(.^(n)C_(2))/(3)+(.^(n)C_(4))/(5)+(.^(n)C_(6))/(7)+"....."+= (2^(n))/(n+1)`

Options

A. `(2^(n+1))/(n+1)`
B. `(2^(n+1)-1)/(n+1)`
C. `(2^(n))/(n+1) `
D. none of these

Answer

c

Explanation

From illustration 14 and 15, we have
`(C_(0))/(1)+(C_(1))/(2) +(C_(2))/(3) +...+(C_(n))/(n+1) = (2^(n+1)-1)/(n+1)`
and `(C_(0))/(1) + (C_(1))/(2) + (C_(2))/(3) - ... +(C_(n))/(n + 1)=(1)/(n+1)`
Adding these two, we get
`2((C_(0))/(1) + (C_(1))/(2) + (C_(2))/(3) +..... )=(2^(n+1))/(n + 1)`
`rArr (C_(0))/(1) + (C_(2))/(3) + (C_(4))/(5) +..... =(2^(n))/(n + 1)`.