Question

Find the sum of the first n terms of the series : `1 + 2(1+1/n) + 3(1+1/n)^2 + 4(1+1/n)^3.......`?

Answer 1

See below.

Explanation:

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

`= d/(dx)((x^(n+2)-1)/(x-1)-1)=(x^(n+1) (n (x-1) + x-2)+1)/(x-1)^2` and making `x = 1+1/n` we get at

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

NOTE:

For `n=3`

`1 + 2 (1 + 1/3) + 3 (1 + 1/3)^2 + 4 (1 + 1/3)^3=499/27`

`3^2 + (1 + 1/3)^3 (1 + 3)=499/27`