Question

S${}_N$ denotes sum of first $n$ terms then find out $\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}$.

Answer 1

Given $S_N$ is the sum of first n natural numbers

So, $S_N=\frac{n(n+1)}{2}$

$\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=\frac{\frac{3n(3n+1)}{2}-\frac{(n-1)n}{2}}{\frac{2n(2n+1)}{2}-\frac{(2n-1)2n}{2}}$

$=\frac{n(4n+3-n+1)}{2n(2n+1-2n+1)}$

$=\frac{n(8n+4)}{2n.2}$

$=2n+1$