Question

If $s_1,s_2,s_3.....s_q$ are the sums of n terms of q A.P.'s whose first terms are 1,2,3.....q and common differences are 1,3,5,....(2q - 1), respectively then $s_1+s_2+s_3....s_q=$

Answer 1

We have $s_1,s_2,s_3.....s_q$ be the sum of first $n$ terms of AP whose first terms are $1,2,3.....q$ and common difference $1,2,3.....(2q-1)$ respectively.

So,

$s_1=\frac{n}{2}(2+(n-1\left)1\right)$

$s_2=\frac{n}{2}(4+(n-1\left)3\right)$

$s_3=\frac{n}{2}(6+(n-1\left)5\right)$

.

.

.

so on...

$s_n=\frac{n}{2}(2q+(n-1\left)\right(2q-1\left)\right)$

$s_1+s_2+.....+s_n=\frac{n}{2}\left[\right(2+4+6+....+2q)+(n-1\left)\right(1+3+5+....+(2q-1)\left)\right]$

$=\frac{n}{2}[\frac{q}{2}\times 2\times (q+1)+(n-1\left)\right(\frac{q}{2}2q\left)\right]$

$=\frac{n}{2}[q^2+q+n{q}^2-q^2]$

$=\frac{nq}{2}[1+nq]$