If ` (1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + C_(3) x^(3) + …+ C_(n) x^(n)` prove that
`(C_(0))/(1) + (C_(2))/(3) + (C_(4))/(5) + ...= (2^(n))/(n+1)` .

Explanation

`because (1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + C_(3) x^(3)+ C_(4) x^(4) + ...+ C_(n) x^(n)` ...(i)
Integrating on both sides of Eq.(i) within limits - 1 to 1
`int_(-1)^(1) (1 + x)^(n) dx = int_(-1)^(1) (C_(0)+ C_(1)x + C_(2)x ^(2) + C_(3)x^(3) + C_(4) x^(4) + ...+ C_(n) x^(n) )dx`
` = int _(-1)^(1) (C_(0) + C_(2) x^(2) + C_(4) x^(4) + ...)dx + int_(-1)^(1) (C_(1) x + C_(3) x^(3) + ...)dx `
` = 2 int_(0)^(1) (C_(0) + C_(2) x^(2) + C_(4)x^(4) + ...)dx + 0 `
[by property of definite integral ]
[since , second integral contians odd funcation ]
`[((1 + x)^(n+1))/(n+1)]_(-1)^(1) = 2 [(C_(0)x + (C_(2)x^(3))/(3) + (C_(4)x^(4))/(5)+ ...)]_(0)^(1) `
or ` C_(0) + (C_(2))/(3) + (C_(4))/(5) + ...+ (2^(n))/(n+1)`
I.Aliter
` LHS = C_(0) + (C_(2))/(3) + (C_(4))/(5) + ...`
` = 1 + (n(n-1))/(1*2*3) + (n(n-1)(n-2)(n-3))/(1*2*3*4*5)+...`
` = (1)/((n+1)) {(n+1)/(1) + ((n+1)n(n-1))/(1*2*3)+ ((n+1)n(n-1)(n-2)(n-3))/(1*2*3*4*5)+...}`
`= (1)/(n+1) {""^(n+1)C_(1) + ""^(n+1)C_(3) + ""^(n+1)C_(5) + ...}`
` = (1)/((n+1))` [sum pf even binomial coeffcient of `(1 + x)^(n+1)]`
` = (2^(n+1 -1))/(n+1) = (2^(n))/(n+1) = RHS `