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Question

The sum to n terms of the series
312+512+22+712+22+32+......... is

A
6nn+1
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B
9nn+1
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C
12nn+1
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D
3nn+1
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Solution

The correct option is A 6nn+1
LetTr=2r+112+22+r2=(2r+1)r(r+1)(2r+1)=6r(2+1)Tr=6r(r+1)=6nr=1(r+1)rr(r+1)=6nr=1[1r1r+1]=6[(112)+(1213)+.......+(1n1n+1)]=6[11n+1]=6nn+1Hence,optionAisthecorrectanswer.

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