# Is there any way to express $\sum_{i=1}^{n} i * 10^{i-1}$ without sum?

I would like to compute the following sum with big values of n :

$$\sum_{i=1}^{n} i * 10^{i-1}$$

I'm wondering if there is a way to express it in a manner that is faster to evaluate.

In other words, can this sum be simplified?

• Compute $\sum_{i=1}^n 10^i$ and differentiate both sides w.r.t. $10$. – StubbornAtom Sep 16 '16 at 12:34
• @StubbornAtom you rather need to differentiate w.r.t. 10 ! – justt Sep 16 '16 at 12:36
• @justt right you are. – StubbornAtom Sep 16 '16 at 12:37
• Thank you for your helpful comment. – Cydonia7 Sep 16 '16 at 12:38

Hint

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

• Thanks, that helped a lot! I did not think of using differentiation there. – Cydonia7 Sep 16 '16 at 12:37

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

$$\sum_{k=0}^n kx^{k-1}=\frac{d}{dx}\frac{x^{n+1}-1}{x-1}=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{(x-1)^2}=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}.$$

• I think there is a sign mistake in the answer :) – Cydonia7 Sep 16 '16 at 12:46
• thanks ! I corrected it :) – Alexandre Krajenbrink Sep 16 '16 at 12:49

This can be a way of expressing it without sum:

\begin{align}\sum_{i=1}^{2} i \times 10^{i-1}&=21\\ \sum_{i=1}^{3} i \times 10^{i-1}&=321\\ \sum_{i=1}^{4} i \times 10^{i-1}&=4321\end{align}\\\vdots\\\sum_{i=1}^{n} i \times 10^{i-1}=n...321