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Question

Find the sum to n terms

12+(12+22)+(12+22+32)+


Solution

The given series is 12+(12+22)+(12+22+32)+ to n terms.

Let an be the nth term of the given series and Sn be the sum of n terms of the given series.

an=(12+22+32++n2)

=n(n+1)(2n+1)6

=16[2n3+3n2+n]

Sn=nk=1ak

=nk=116[2k3+3k2+k]

=16[2.13+3.12+1]+16[2.23+3.22+2]++16[2n3+3n2+n]

=16[2(13+23++n3)+3(12+22++n2)+(1+2++n)]

=16[2.{n(n+1)2}+3n(n+1)(2n+1)6+n(n+1)2]

=16[n2(n+1)22+n(n+1)(2n+1)2+n(n+1)2]

=n(n+1)12[n(n+1)+2n+1+11]

=n(n+1)12(n2+3n+2)

=n(n+1)(n+1)(n+2)12

=n(n+1)2(n+2)12


Mathematics

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