The sum to 50 terms of the series 312+512+22+712+22+32+… is
A
10017
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B
15017
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C
20051
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D
5017
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Solution
The correct option is A10017 Let Tr be the rth term of the given series. Then, Tr=2r+112+22+……+r2 =6(2r+1)(r)(r+1)(2r+1)=6r(r+1) =6(1r−1r+1)
So, sum is given by 50∑r=1Tr=650∑r=1(1r−1r+1) =6[(1−12)+(12−13)+(13−14)+……+(150−151)] =6[1−151] =10017