Question

Prove : $1+2+3+...+n=\frac{1}{2}n(n+1)$

Answer 1

The series $1+2+3+.......+n$ is in A.P.

where

first term, $a=1$

common difference, $d=1$

last term, $l=n$

No. of terms $=n$

In A.P.

Sum of first $n$ terms, $S_n=\frac{n}{2}(a+l)$

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