Question

Find the sum of first $n$ natural numbers.

• A. $\frac{n(2n+1)}{2}$
• B. $\frac{2n+1}{2}$
• C. $\frac{n(n+1)}{2}$
• D. $\frac{n(2n-1)}{2}$

Answer 1

The correct option is C

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

The first $n$ natural numbers are 1, 2, 3, 4,....., $n.$
This sequence forms an AP, whose first term is 1 and the common difference is 1.

The sum upto $n$ terms of an AP is given by
$S_n=\frac{n}{2}[2a+(n-1\left)d\right].$

Hence, the sum of the above AP is given by

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

$\Rightarrow S_n=\frac{n}{2}[1+n]$

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