Write the sum of first n even natural numbers.

In this problem, we need to find the sum of first n even natural numbers.

So, we know that the first odd natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 2

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

So, for n terms,

$S_{n}=\frac{n}{2}[2(2)+(n-1) 2]$

$=\frac{n}{2}[4+2 n-2]$

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

$=n(1+n)$

Therefore, the sum of first $n$ even natural numbers is $S_{n}=n(n+1)$.