Question

If the sum of the first ten terms of the series $\bigg \lgroup1\frac{3}{5}\bigg \rgroup^{2}+\bigg \lgroup2\frac{2}{5}\bigg \rgroup^{2}+\bigg \lgroup 3 \frac{1}{5}\bigg \rgroup^{2}+4^{2}+\bigg \lgroup 4\frac{4}{5}\bigg \rgroup^{2}+......\ is\ \frac{16}{5}m$ , then m is equal to :
(a) 100
(b) 99
(c) 102
(d) 101

Answer 1

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