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Question

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 A 10017
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(1r1r+1)
So, sum is given by
50r=1Tr=650r=1(1r1r+1)
=6[(112)+(1213)+(1314)++(150151)]
=6[1151]
=10017

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