Question:
Find the sum of first n even natural numbers.
Solution:
The first n even natural numbers are 2 ,4, 6, 8, 10, ..., n.
Here, a = 2 and d = (4 - 2) = 2
Sum of $n$ terms of an AP is given by
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\left(\frac{n}{2}\right) \times[2 \times 2+(n-1) \times 2]$
$=\left(\frac{n}{2}\right) \times[4+2 n-2]=\left(\frac{n}{2}\right) \times(2 n+2)=n(n+1)$
Hence, the required sum is n(n+1).