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

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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 = [tex] \sf{\dfrac{[2a+(n-1)d] + [2a+(n-1)d] + … + [2a+(n-1)d] + [2a+(n-1)d]}{n~times}}[/tex]

   OR ,

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

   OR ,

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

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

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

   Sum of squares of ‘n‘ natural numbers

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

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