Ideas for a Closed form for $ \sum_{k=0}^n k10^k$

Question

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.

Answer 1

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?

Answer 2

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?

Answer 3

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

$$\sum_{k=1}^n k 10^k = \frac{d}{dx}\left(\frac{x^{n+1}-1}{x-1}\right) (10)$$

Answer 4

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.