Question

find the sum of squares of 'n' natural number (give brief ansanswer ​

Answer 1

Sum of 'n' natural numbers

Let the sum of the first n terms of AP :

a , a + d , a + 2d ...

The n th term for this AP is a+(n-1)d . Let S denote the sum of the first n terms of the AP . We have

S = a+(n-1)d + (a+2d) + ... + [a+(n-1)d] ___ ( 1 )

Rewriting the terms in reverse order , we have

S = [a+(n-1)d] + [a+(n-2)d] + ... + (a+d) + a ___ (2)

On adding ( 1 ) & (2) term - wise . We get

2S = $\sf{\dfrac{[2a+(n-1)d] + [2a+(n-1)d] + ... + [2a+(n-1)d] + [2a+(n-1)d]}{n~times}}$

OR ,

2S = n [ 2a + (n-1)d ] ( since there are n terms )

OR ,

S = $\sf{\dfrac{n}{2}} [2a+(n-1)d]$

So , the sum of the first n terms of an AP is given by

S = ${\sf{\bold{\dfrac{n}{2}}[2a+(n-1)d]}}$

Sum of squares of 'n' natural numbers

Σn² = $\sf {\dfrac{[n(n+1)(2n+1)]}{6}}$