Question:
If the sum of the first ten terms of the series $\left(1 \frac{3}{5}\right)^{2}+\left(2 \frac{2}{5}\right)^{2}+\left(3 \frac{1}{5}\right)^{2}+4^{2}+\left(4 \frac{4}{5}\right)^{2}+\ldots$, is $\frac{16}{5} \mathrm{~m}$, then $\mathrm{m}$ is equal to :-
Correct Option: , 3
Solution:
