If `(1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3)x^(3)`
` + ...+ C_(n)x^(n)` , prove that ` (C_(1))/(2) + (C_(3))/(4) + (C_(5))/(6) + …= (2^(n)-1)/(n+1)` .

Explanation

We know that , from Example
`C_(0) + (C_(1))/(2) + (C_(2))/(3) + (C_(3))/(4) + (C_(4))/(5) + (C_(5))/(6) + ...= (2^(n+1) -1)/(n+1)` ...(i)
and `C_(0) + (C_(1))/(2) + (C_(2))/(3) + (C_(3))/(4) + (C_(4))/(5) + (C_(5))/(6) + ...= (1)/(n+1)` ...(ii)
On subtracting Eq.(ii) from Eq.(i) , we get
`2((C_(1))/(2) + (C_(3))/(4) - (C_(3))/(5)+ ...) = (2^(n+1) -2)/(n+1)`
On dividing each sides by 2, we get
`(C_(1))/(2) + (C_(3))/(4) - (C_(3))/(5)+ ... = (2^(n)-1)/(n+1)`
I. Aliter ` LHS = (C_(1))/(2) + (C_(3))/(4) + (C_(5))/(6) + ...`
` = (n)/(1*2) + (n(n-1)(n-2))/(1*2*3*4) `
` + (n(n-1)(n-2)(n-3)(n-4))/(1*2*3*4*5*6) + ...`
`= (1)/(n+1) [((n+1)n)/(1*2) + ((n+1)n(n-1)(n-2))/(1*2*3*4) + ((n+1)n(n-1)(n-2) (n-3)(n-4))/(1*2*3*4*5*6)+...]`
Put n+ 1 = N , then
`LHS = (1)/(N)[(N(N-1))/(2!)+ (N(N-1)(N-2)(N-3))/(4!)+ (N(N-1)(N-2)(N-3)(N-4)(N-5))/(6!)+...]`
`= (1)/(N) [""^(N)C_(2)+ ""^(N)C_(4) + ""^(N)C_(6)+ ...]`
`= (1)/(N)[(""^(N)C_(0) + ""^(N)C_(2) + ""^(N)C_(4)+ ""^(N)C_(6)+ ...)- ""^(N)C_(0)]`
` = (1)/(N) [ 2^(N-1) -1] = (2^(n)-1)/(n+1) = RHS`
II.Aliter
`LHS = (C_(1))/(2) + (C_(3))/(4) + (C_(5))/(6) + ...`
Case I If n is odd say ` n = 2m + 1 ,AA m iin W `, then
`LHS = sum_(r=1)^(m) (""^(2m+1)C_(2r+1))/(2r + 2) = sum_(r=0)^(m)(""^(2m+2)C_(2r+2))/(2m+2) " "[because (""^(2m+2)C_(2r+2))/(2m+1) = (""^(2m+1)C_(2r+1))/(2r + 2)]`
`= (1)/((2m+2)) (""^(2m+2)C_(2) + ""^(2m+2)C_(4) + ...+ ""^(2m+2)C_(2m +2))`
` = (1)/((2m+2)) * (2^(2m +2-1" _ "2m +1)C_(0)) = (2^(n) -1)/(n+1)" " [because 2m + 1 = n]` = RHS
Case II If n is even say ` n = 2m, AA m in N` , then
`LHS = sum_(r=0)^(m-1) (""^(2m )C_(2r +1))/((2r +2)) = sum_(r=0)^(m-1) (""^(2m+1 )C_(2r +2))/((2r +1))" " [because (""^(2m+1 )C_(2r +2))/(2m +1)= (""^(2m )C_(2r +1))/(2m +2)]`
`= (1)/((2m+1)) sum_(r=0)^(m-1) ""^(2m+1)C_(2r +2)`
`= (1)/((2m +1))(""^(2m+1)C_(2) + ""^(2m+1)C_(4) + ""^(2m+1)C_(6) + ...+""^(2m+1)C_(2n))`
`= (1)/((2m +1))*(2^(2m +1-1" - "2m+1)C_(0)) `
` (2^(n) -1)/(n+1) = RHS " " [ because n = 2 m] `