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Question

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

A
112n(n+1)(2n3+3n2+n)
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B
112n(n+1)(3n3+2n2+n)
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C
16n(n+1)(2n+1)
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D
14n2(n+1)2
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Solution

The correct option is A 112n(n+1)(2n3+3n2+n)
nth term Tn=12+22+32+.....+n2=n2=16n(n+1)(2n+1)=16(2n3+3n2+n)
Thus required summation is =Tn=16(2n3+3n2+n)
=16(2n2(n+1)24+3n(n+1)(2n+1)6+n(n+1)2)=112n(n+1)(2n3+3n2+n)

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