Summation of a series I'm trying to solve a recurrence relation and came across this term $\sum_{i=0}^n i9^i$? I thought this was a geometric series, but I guess it's not. Is it possible to solve this?
 A: HINT: The geometric series you refer to has closed form:
$$\frac{1-x^{n+1}}{1-x} = \sum_{k=0}^{n}x^k$$
Can you figure out through differentiation how to derive the solution for $\sum_{k=0}^{n}kx^k$? Then plugging in $9$ will give you the particular solution you are looking for.
EDIT
The derivative of the RHS above is
$\sum_{k=0}^{n}kx^{k-1}$.  After multiplying by $x$ we get the general form of your series in question.
Next we differentiate the LHS and get: 
$$\frac{-(n+1)x^n(1-x)-(-1)(1-x^{n+1})}{(1-x)^2}$$
$$=\frac{1-x^{n+1}-nx^n+nx^{n+1}-x^n+x^{n+1}}{(1-x)^2}$$
$$=\frac{1-x^n(n+1)+nx^{n+1}}{(1-x)^2}$$
After multiplying by $x$ we have finally:
$$\sum_{k=0}^{n}kx^k = \frac{x-x^{n+1}(n+1)+nx^{n+2}}{(1-x)^2}$$
A: Let $S$ be our sum. Then 
$$S=9+2\cdot 9^2+3\cdot 9^3+\cdots+n\cdot 9^n.$$
Multiply through by $9$. We get
$$9S=9^2+2\cdot 9^3+3\cdot 9^4+\cdots+n\cdot 9^{n+1}.$$
Subtract. We get
$$8S=-(9+9^2+9^3+\cdots +9^n)+n\cdot 9^{n+1}.$$
You probably know how to find a closed form for the geometric series $9+9^2+9^3+\cdots+9^n$. Now simplify, and solve for $S$.
