Find the sum of first and even natural numbers.

To Do: Find the sum of first n even natural numbers

Solution:

The n even no. will be in the form; $2+4+6+...+2n$

We have to solve using Sum of n terms of an Arithmetic Progression formula.

We know Sum of n terms of an Arithmetic Progression = $\frac{n(2a+(n-1)d)}{2}$

Here, $a = 2, n = n, d = 2$

So, $2+4+6+...+2n = \frac{n(2\times2 +(n-1)2}{2}$

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

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

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

= $n(n+1)$

So, sum of first n even natural numbers $n(n+1)$

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Updated on: 10-Oct-2022

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