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JEEMAIN and NEET
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JEEMAIN PAST PAPERS
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2016
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If the sum of the first ten terms of the series $(1 + \frac{3}{5})^2 + ( 2 + \frac{2}{5})^2 + (3 + \frac{1}{5})^2 + 4^2 + (4 + \frac{4}{5})^2......,$ is $\frac{16}{5} m$, then $m$ is equal to :
( A ) 99
( B ) 101
( C ) 102
( D ) 100
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