# tricky summation proof by induction

I need to prove the following identity by induction: $$\sum_{i=0}^ni\cdot n^{i-1}=(n-1)\cdot 2^n+1$$

I have the base case where $$n=1$$ and all that, but I'm stuck on how to turn the inductive step into the final solution. I know I need to assume the identity holds for $$n-1$$, but I'm stuck on how to use that to show it holds true for n.

• 1. Did you already check this equation for some $n$? – miracle173 Feb 26 at 10:20
• 2. Why do you want to proof it with induction? – miracle173 Feb 26 at 10:22
• Note: use ^{i-1} instead of ^(i-1) to get the expected display. – Arnaud Mortier Feb 26 at 10:25
• This cannot be true. The right hand side include a term of $n\cdot n^{n-1}=n^n$ which is $>> n\cdot 2^n$ – Thomas Lesgourgues Feb 26 at 11:04

## 1 Answer

Your identity seems wrong to me. I took $$n=5$$ then the L.H.S. becomes, $$\displaystyle \sum_{i=0}^5 i\cdot 5^{i-1}=0+1+10+75+500+3125=3711$$ but R.H.S. becomes $$(5-1)\cdot 2^5+1=2^7+1=129$$ whis is absurd!

Note: However, putting the LHS on Mathematica gives $$\sum_{i=0}^ni\cdot n^{i-1}=\frac{1-n^n-n^{1+n}+n^{2+n}}{(n-1)^2}$$

Hoping for this to be true for all $$n\in\mathbb{N}\setminus\{0\}$$.