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

Solution :
$$\eqalign{ & {\left( {\frac{8}{5}} \right)^2} + {\left( {\frac{{12}}{5}} \right)^2} + {\left( {\frac{{16}}{5}} \right)^2} + {\left( {\frac{{20}}{5}} \right)^2}..... + {\left( {\frac{{44}}{5}} \right)^2} \cr & S = \frac{{16}}{{25}}\left( {{2^2} + {3^2} + {4^2} + ..... + {{11}^2}} \right) \cr & = \frac{{16}}{{25}}\left( {\frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1} \right) \cr & = \frac{{16}}{{25}} \times 505 = \frac{{16}}{5} \times 101 \cr & \Rightarrow \frac{{16}}{5}m = \frac{{16}}{5} \times 101 \cr & \Rightarrow m = 101. \cr} $$