Question

If $S_1,S_2,S_3...S_p$ be the sum of $n$ terms of '$p$' A.P.'s, whose first terms are respectively $1,2,3$... and common difference are respectively, $1,2,3$ ... Prove that :
$S_1+S_2+S_3+...+S+p=\frac{np}{4}(n+1)(p+1)$

Answer 1

For the first A.P.,

first term, $a=1$

common difference, $d=1$

Sum of $n$ terms, $=\frac{n}{2}\{2a+(n-1\left)d\right\}$

$S_1=\frac{n}{2}\left\{2\right(1)+(n-1\left)\right(1\left)\right\}$

$S_1=\frac{n}{2}\{n+1\}$

For the second A.P.,

first term, $a=2$

common difference, $d=2$

Sum of $n$ terms, $=\frac{n}{2}\left\{2\right(2)+(n-1\left)\right(2\left)\right\}$

$S_2=\frac{n}{2}\{2n+2\}$

$S_2=2\frac{n}{2}\{n+1\}$

Similarly,

$S_3=3\frac{n}{2}\{n+1\}$

..

..

$S_p=p\frac{n}{2}\{n+1\}$

$S_1+S_2+S_3+......+S_p=\frac{n}{2}(n+1)(1+2+3+....+p)$

$=\frac{n}{2}(n+1)\frac{p}{2}(p+1)$

$=\frac{np}{4}(n+1)(p+1)$