Recent content by vsector

Oh! I see. So that was what he meant. Thanks.

Brilliant! Lesse, S = a + 2a^2 + 3a^3 + \ldots + na^n aS = a^2 + 2a^3 = 3a^4 + \ldots + na^{n+1} Subtracting, I get: S(1 - a) = a + a^2 + a^3 + a^4 + \ldots + a^n + na^{n + 1} S(1 - a) = \frac{a^{n+1} - 1}{a - 1} - 1 + na^{n + 1} S = \frac{a^{n+1} - 1}{(a - 1)(1 - a)} - \frac{1}{1 - a} +...

I'm sorry, I'm really lost. There's a relation between summation and integration somewhere that I can use to my advantage is what I got from your reply. It's been a while since I've done any Calculus. Is there a resource available that I can study this more in depth? All I have is an...

Homework Statement Evaluate the sums: \sum^{n}_{i=1}ia^{i} Homework Equations The Attempt at a Solution I'm assuming that because there's no limit or anything, the professor wants an equation. I know \sum^{n}_{i=1}i is \frac{1}{2}n(n+1) and \sum^{n}_{i=1}a^{i} is \frac{a^{n+1} - 1}{a-1}-1...