Question

If $S_1,S_2,....S_n$ 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. $S_1+S_2+2S_3+3S_4+....(n-1)$

$S_n=1^n+2^n+3^n+...+n^n$.

Answer 1

Given,

$S_1,S_2,.....,S_n$ are the same of n terms of an G.P. whose first term is 1 in each case and the common ratios are 1, 2, 3, ...., n.

$\therefore S_1=1+1+1+.....$ n terms = n .....(i)

$S_2=\frac{1(2^n-1)}{2-1}=2^n-1$ ...........(ii)

$S_3=\frac{1(3^n-1)}{3-1}=\frac{3^n-1}{2}$ ........ (iii)

$S_4=\frac{1(4^n-1)}{4-1}=\frac{4^n-1}{3}$ ....... (iv)

$S_n=\frac{1(n^n-1)}{n-1}=\frac{n^n-1}{n-1}$ ......(n)

Now, LHS = $S_1+S_2+2S_3+3S_4+....+(n-1)S_n$

$=n+2^n-1+3^n-1+4^n-1+....+n^n$ [Using (1), (2), (3), ....., (n)]

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

$=n+(2^n+3^n+4^n+....+n^n)-(n-1)$

$=n+(2^n+3^n+4^n+....+n^n)-n+1$

$=1+2^n+3^n+4^n+....+n^n$

$=1^n+2^n+3^n+4^n+....+n^n$

= RHS

Hence proved.