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Question

Find the sum of 12+(12+22)+(12+22+32)+.......

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Solution

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(2nn=1n3+3nn=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


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