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