If the sum of the first ten terms of the series

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 :-

  1. 99

  2. 102

  3. 101

  4. 100


Correct Option: , 3

Solution:

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