Question

If `(1+x)^n=c_0+c_1x+c_2x^2+cdots+c_nx^n` then show that `c_0+3c_1+5c_2+cdots+(2n+1)c_n=(n+1)2^n`?

Did you mean:
If `(1+x)^n =C_0+C_1x+C_2x^2+cdots+C_nx^n` then show that `C_0+3C_1+5C_2+cdots+ (2n+1)C_n=(n+1)2^n`

Answer 1

See below.

Explanation:

Taking now

`2d/(dx)(1+x)^n + (1+x)^n = 2n(1-x)^(n-1)+(1+x)^n = (2n+(1+x))(1+x)^(n-1)`

we have also

`2d/(dx)(1+x)^n + (1+x)^n = 2 sum_(k=0)^n k c_k x^(k-1) + sum_(k=0)^n c_k x^k =`

and making now `x = 1` we have

`sum_(k=0)^n (2k+1)c_k = (2n+(1+1))(1+1)^(n-1) = (n+1)2^n`