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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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