Ideas for a Closed form for $ \sum_{k=0}^n k10^k$ Is there a closed formula for this summation:
$$ \sum_{k=0}^n k10^k, $$
where $n\in\mathbb{N}$? I would like to learn trick o strategies for this kind of problems.
 A: HINT
\begin{align*}
\sum_{k=0}^{n}k10^{k} = \sum_{k=0}^{n}(k+1)10^{k} - \sum_{k=0}^{n}10^{k}
\end{align*}
EDIT
In order to obtain the desired result, notice that
\begin{align*}
s_{n}(x) = 1 + x + x^{2} + x^{3} + \ldots + x^{n} & \Rightarrow s_{n}(x)x = x + x^{3} + \ldots + x^{n+1}\\\\
& \Rightarrow s_{n}(x) - s_{n}(x)x = 1 - x^{n+1}\\\\
& \Rightarrow s_{n}(x)(1 - x) = 1 - x^{n+1}\\\\
& \Rightarrow s_{n}(x) = \frac{1 - x^{n+1}}{1 - x}
\end{align*}
Can you take it from here?
A: You can change the summation to start from $k=1$. Then
$$ s\cdot \frac{d}{ds} \sum\limits_{k=1}^{n} s^k = s \cdot \sum\limits_{k=1}^n k s^{k-1} =\sum\limits_{k=0}^n ks^k. $$
Now, the left hand side has a nice closed expression that we can then evaluate at $s=10$, can you find the closed expression?
A: For $x \gt 0$, you have
$$f_n(x)=\sum_{k=1}^n x^k = \frac{x^{n+1}-1}{x-1} - 1$$ hence
$$x f^\prime_n(x) = \sum_{k=1}^n k x^k$$ and therefore
$$\sum_{k=1}^n k 10^k = \frac{d}{dx}\left(\frac{x^{n+1}-1}{x-1}\right) (10)$$
A: Many answers have used the derivative method, but you can also do it this way:
Let $S$ be your required summation. Write $10S$ and subtract from $S$. You would get a GP series which is easily evaluated.
