Find the sum to n terms
12+(12+22)+(12+22+32)+…
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