Closed form solution for series

[EvLer]

I am kind of blocked on how to find closed form solution for these finite series:

sigma [naⁿ]
(n = 0 to k)
k = any integer

thanks for any hints... and i can't find LaTex tutorials, where are they?

 

[benorin]

So you may not know how to sum

$$\sum_{n=0}^{k} na^n$$

so what. You do know how to sum something similar, right? like, say,

$$\sum_{n=0}^{k} a^n$$

and is not $$\frac{d}{da}a^n = na^{n-1}$$?

so figure it out from there. (left click one of the expressions to see how to write them with symbols.)

 

[benorin]

A new sum from an old sum

To sum $$\sum_{n=0}^{k} na^n,$$

recall the sum of something similar, like, say,

$$\sum_{n=0}^{k} a^n = \frac{1-a^{k+1}}{1-a}$$

and consder that $$\frac{d}{da}a^n = na^{n-1}$$

that one may write

$$\frac{d}{da}\left( \sum_{n=0}^{k} a^n\right) = \frac{d}{da}\left( \frac{1-a^{k+1}}{1-a}\right)\Rightarrow \sum_{n=0}^{k}\left( \frac{d}{da}a^n\right) = \sum_{n=1}^{k}na^{n-1} =\frac{-(k+1)a^{k}(1-a)-(1-a^{k+1})(-1)}{(1-a)^2}$$

from the last equality, we have

$$a\sum_{n=1}^{k}na^{n-1} =a\frac{-(k+1)a^{k}(1-a)+(1-a^{k+1})}{(1-a)^2}\Rightarrow \sum_{n=1}^{k}na^{n} =\frac{a}{1-a}\left[ \frac{1-a^{k+1}}{1-a}-(k+1)a^{k}\right]$$

and since $$0+\sum_{n=1}^{k}na^{n} = \sum_{n=0}^{k}na^{n},$$

we have

$$\boxed{ \sum_{n=0}^{k}na^{n} = \frac{a}{1-a}\left[ \frac{1-a^{k+1}}{1-a}-(k+1)a^{k}\right] = \frac{a}{(1-a)^2}\left[ 1+ka^{k+1}-(k+1)a^{k} \right] }$$​