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Question

1+4+10+19+...3n23n+22=

A
n2(n2+1)2
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B
n(n2+1)2
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C
n2(n+1)2
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D
{n(n+1)2}2
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Solution

The correct option is D n(n2+1)2
nk=13k23k+22=32nk=1k232nk=1k+nk=11

=32n(n+1)(2n+1)632n(n+1)2+n

=n(n21)2+n

=n(n2+1)2

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