Problem:
$S_1, S_2, S_3, \ldots ,S_n$ are the sum of $n$ terms of $n$ GPs whose first term is 1 in each case.
However, the common ratios $r$ are $1,2, 3, \ldots,n$ respectively.
Prove that $$S_1 + S_2 + 2 S_3 + 3 S_4 +\cdots+ (n - 1) S_n = 1^n + 2^n + 3^n + \cdots + n^n $$
My attempt:
$$S_1=0$$ $$S_2=(2^n-1)$$ $$2S_3=(3^n-1)$$ $$\vdots$$
$$(n-1)S_n=(n^n-1)$$ $$\Longrightarrow S_1 + S_2 + 2 S_3 + 3 S_4 +\cdots+ (n – 1) S_n =2^n+3^n+4^n+\cdots+n^n - (n-1)$$ Unfortunately I am unable to proceed any further. I would be truly grateful if somebody would please show me how to proceed. Many, many thanks!
Note that $S_1=n$, not $0$. $a+ar+ar^2+\cdots+ar^{n-1}=\frac{a(r^n-1)}{r-1}$ is true for $r\not=1$.
We have $S_1=1+1+\cdots+1=n$ (because it has $n$ terms)
For $i\ge 2$, we have $$S_i=1+i+i^2+\cdots+i^{n-1}=\frac{i^n-1}{i-1}.$$ Hence, we have $$\begin{align}S_1+\sum_{i=2}^{n}(i-1)S_i&=n+\sum_{i=2}^{n}(i-1)\cdot\frac{i^n-1}{i-1}\\&=n+\sum_{i=2}^{n}(i^n-1)\\&=n-(n-2+1)+\sum_{i=2}^{n}i^n\end{align}$$