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 A6nn+1 LetTr=2r+112+22+r2=(2r+1)r(r+1)(2r+1)=6r(2+1)∑Tr=∑6r(r+1)=6∑nr=1(r+1)−rr(r+1)=6∑nr=1[1r−1r+1]=6[(1−12)+(12−13)+.......+(1n−1n+1)]=6[1−1n+1]=6nn+1Hence,optionAisthecorrectanswer.