Find the sum of 12+(12+22)+(12+22+32)+.......
First term=12
Second term=12+22
Third term=12+22+32...
nth term=12+22+32+...+n2
Here an=12+22+32+...+n2
=n(n+1)(2n+1)6
=2n3+n2+2n2+n6
=2n3+3n2+n6
Now, the sum of n terms is
Sn=∑nn=1an
=∑nn=12n3+3n2+n6
=16(2∑nn=1n3+3∑nn=1n2+∑nn=1n)
=16(2n2(n+1)22+3n(n+1)(2n+1)6+n(n+1)2)
=n(n+1)12[n(n+1)+(2n+1)+1]
=n(n+1)12[n2+n+2n+1+1]
=n(n+1)12[n2+3n+2]
=n(n+1)12[n2+2n+n+2]
=n(n+1)12[(n+1)(n+2)]
=n(n+1)2(n+2)12
Thus, the required sum is n(n+1)2(n+2)12