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Question

Find the sum of n terms of 12+(12+22)+(12+22+32)+(12+22+32+42)+... from that find the sum of the first 10 terms

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Solution

12+(12+22)+(12+22+32)+..........n terms

=ni=1(12+22+.....i2)

=ni=1ij=1j2

=ni=1i(i+1)(2i+1)6

=ni=12i3+3i2+i6

=ni=113i3+nj=112j2+nk=116k

=13(n2(n+1)24)+12(n(n+1)(2n+1)6)+16(n(n+1)2)

=n2(n+1)212+n(n+1)(2n+1)12+n(n+1)12

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

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

S10=10(10+1)2(10+2)12

=10(11)2(12)12

=1210

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