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] Reply

Sumof‘n’naturalnumbersLet the sum of the first

terms of AP :n,aa+,da+2d…The

th term for this AP isna+(n-1). LetdSdenote the sum of the firstterms of the AP . We havenS=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)+___ (2)aOn adding ( 1 ) & (2) term – wise . We get

2S## OR ,

2Sn[2a+(n-1)d]( since there areterms )n## OR ,

S=So ,

thesumofthefirstntermsofanAPisgivenbyS=Sumofsquaresof‘n‘naturalnumbersΣn²= [tex] \sf {\dfrac{[n(n+1)(2n+1)]}{6}} [/tex]