First slide

Special Series in Sequences and Series

Question

If the sum of the first ten terms of the series ${(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + \dots . is \frac{16}{5} m, then m is equal to$

Moderate

A

99
B

102
C

101
D

100

Solution

${(\frac{8}{5})}^{2} + {(\frac{12}{5})}^{2} + {(\frac{16}{5})}^{2} + {(\frac{21}{5})}^{2} + {(\frac{24}{5})}^{2} + ....$
$\begin{array}{l} S = \frac{4^{2}}{25} [1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots 11^{2} - 1] \\ = \frac{16}{25} [\frac{(11) (12) (23)}{6} - 1] \\ = \frac{16}{25} [505] = \frac{16}{5} m \\ m = 101 \end{array}$

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