If the sum of the first ten terms of the series \(\left (1 \dfrac {3}{5}\right )^{2} + \left (2 \dfrac {2}{5}\right )^{2} + \left (3 \dfrac {1}{5}\right )^{2} + 4^{2} + \left (4 \dfrac {4}{5}\right )^{2} + ..............\) is \(\dfrac {16}{5}m\), then \(m\) is equal to:
1
\(102\)
2
\(101\)
3
\(100\)
4
\(99\)