Question

IF `(1+x)^n=c_0+c_1x+c_2x^2+cdots+c_nx^n` then show that `c_1+2c_2+3c_3+cdots+nc_n=n.2^(n-1)`?

Answer 1

See below.

Explanation:

We have

`d/(dx)(1-x)^n = n(1+x)^(n-1) = c_1+2c_2x+3c_3x^2+ cdots+n c_n x^(n-1)`

Making now `x = 1` we get at

`n(1+1)^(n-1) = c_1+2c_2+3c_3+ cdots+n c_n`

or

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