How to apply property to a combined sum? I know I don't have the vocabulary to describe this problem.  
The easiest way to describe is by showing.
I have the following sum from a recursion:
$$
\sum_{i = 1}^k{a^{k-i}b^i*i}
$$
And the following two properties:
$$
\sum_{i=1}^k{a^{k-i}} = \frac{x^k-1}{x-1}\\
\sum_{i=1}^k{i*b^i} = \frac{b-(k+1)b^{k+1} + k*b^{k+2}}{1-b^2}\\
$$
How would I go about applying the two properties here?  Is it possible?
 A: Hint. One may just write
$$
\sum_{i = 1}^k{a^{k-i}b^i\cdot i}=a^k\sum_{i = 1}^ki\cdot{\left(\frac{b}{a}\right)^i}, \quad a\neq b.
$$
A: Given that:
$$\sum_{n=1}^knb^i=\frac{b-(k+1)b^{k+1}+kb^{k+2}}{1-b^2}$$
let $b\to\frac ba=ba^{-1}$
$$\sum_{n=1}^knb^na^{-n}=\frac{ba^{-1}-(k+1)b^{k+1}a^{-k-1}+kb^{k+2}a^{-k-2}}{1-b^2a^{-2}}$$
Then multiply both sides by $a^k$
$$a^k\sum_{n=1}^knb^na^{-n}=\sum_{n=1}^knb^na^{k-n}=a^k\left(\frac{ba^{-1}-(k+1)b^{k+1}a^{-k-1}+kb^{k+2}a^{-k-2}}{1-b^2a^{-2}}\right)$$
A: Just as Olivier Oloa answered $$\sum_{i = 1}^k{a^{k-i}b^i\, i}=a^k\sum_{i = 1}^ki\,{\left(\frac{b}{a}\right)^i}=a^k\frac{b}{a}\sum_{i = 1}^ki\,{\left(\frac{b}{a}\right)^{i-1}}={a^{k-1}}b\left(\sum_{i = 1}^k{\left(\frac{b}{a}\right)^{i}}\right)'$$ Assuming $b\neq a$, we just need to consider (using $x=\frac b a$) $$\sum_{i = 1}^k x^i=\frac{x \left(x^k-1\right)}{x-1}\implies\left(\sum_{i = 1}^k x^i\right)'=\frac{(k (x-1)-1) x^k+1}{(x-1)^2}$$ Replacing $x$ and simplifying, then
$$\sum_{i = 1}^k{a^{k-i}b^i\, i}=\frac{b \left(a^{k+1}-b^k (a k+a-b k)\right)}{(a-b)^2}$$
