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1/2 [ 2{n(n+1)/2}^2 – n(n+1)(2n+1)/6 + n(n+1)/2 ] = 1/12 [ n(n+1) {3(n^2) + n +2}]Please show how??
taking n(n+1)/6 common, we get1/2 [ 2{n(n+1)/2}^2 – n(n+1)(2n+1)/6 + n(n+1)/2]=[3n(n+1) – (2n+1)+3]*n(n+1)/12=[3n^2+3n – 2n – 1+3]*n(n+1)/12=1/12 [ n(n+1) {3(n^2) + n +2}]
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