$1+(1+3)+(1+3+5)+(1+3+5+7)+\ldots$ to $10$ terms $=$

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

The sum of the series $1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$ is

$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 < x < \pi, x \neq \frac{\pi}{2}$,then $x=$

For any odd integer $n \ge 1$,${n^3} - {(n - 1)^3} + \dots + {( - 1)^{n - 1}}{1^3} = $

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $a_n = n^2 + 2^n$.