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.......?

1 Answer
Aug 27, 2017

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