If the sum of the first ten terms of the series +…, is , then m is equal to

Engineering

Mathematics

Special Types of Sequence

Question

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 $\frac{16 m}{5}$ , then m is equal to

101

102

100

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$T_{n} = {(\frac{4 n + 4}{5})}^{2} = \frac{16}{25} {(n + 1)}^{2}$

$S_{10} = \frac{16}{25}$ (2² + 3² + ……. + 11²)

$= \frac{16}{25} (\frac{11 \cdot 12 \cdot 23}{6} - 1) = \frac{16}{25} (505)$

$= \frac{16}{5} . (101) .$