# Solve the following

Question:

If S1S2, ..., Sn are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S1 + S2 + 2S3 + 3S4 + ... (n − 1) Sn = 1n + 2n + 3n + ... + nn.

Solution:

Given:

$S_{1}, S_{2}, \ldots, S_{n}$ are the sum of $n$ terms of an G.P. whose first term is 1 in each case and the common ratios are $1,2,3, \ldots, n$.

$\therefore S_{1}=1+1+1+\ldots n$ terms $=n$   ...(1)

$S_{2}=\frac{1\left(2^{n}-1\right)}{2-1}=2^{n}-1$   ...(2)

$S_{3}=\frac{1\left(3^{n}-1\right)}{3-1}=\frac{3^{n}-1}{2}$   ...(3)

$S_{4}=\frac{1\left(4^{n}-1\right)}{4-1}=\frac{4^{n}-1}{3}$   ...(4)

$S_{n}=\frac{1\left(n^{n}-1\right)}{n-1}=\frac{n^{n}-1}{n-1} \quad \ldots \ldots \ldots(\mathrm{n})$

Now, LHS $=S_{1}+S_{2}+2 S_{3}+3 S_{4}+\ldots+(n-1) S_{n}$

$=n+2^{n}-1+3^{n}-1+4^{n}-1+\ldots+n^{n}-1 \quad[$ Using $(1),(2),(3), \ldots,(n)]$

$=n+\left(2^{n}+3^{n}+4^{n}+\ldots+n^{n}\right)-[1+1+1+\ldots+(n-1)$ times $]$

$=n+\left(2^{n}+3^{n}+4^{n}+\ldots+n^{n}\right)-(n-1)$

$=n+\left(2^{n}+3^{n}+4^{n}+\ldots+n^{n}\right)-n+1$

$=1+2^{n}+3^{n}+4^{n}+\ldots+n^{n}$

$=1^{n}+2^{n}+3^{n}+4^{n}+\ldots+n^{n}$

= RHS

Hence proved.