If the sum of the first ten terms of the series 1 3 5 2 + 2 2 5 2 + 3 1 5 2 + 4 2 + 4 4 5 2 + … . , is…

${\left(\frac{8}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2+{\left(\frac{16}{5}\right)}^2$ ........ upto ten terms

$T_n={\left(\frac{4n+4}{5}\right)}^2$

$T_n=16\left(\frac{n^2+2n+1}{25}\right)$

$T_n=\frac{16}{25}\left(n^2+2n+1\right)$

$S_n=\frac{16}{25}\sum _{}\left(n^2+2n+1\right)$

$S_n=\left(\frac{16}{25}\right) \left(\frac{n\left(n+1\right) \left(2n+1\right)}{6}+\frac{2\left(n\right)\left(n+1\right)}{2}+n\right)$

Put $n=10$

$\frac{16}{25}\left(\left(\frac{10·11·21}{6}\right)+\frac{2·10·11}{2}+10\right)$

$=\frac{16}{25} \left(385+110+10\right)=\frac{16}{25}· 505$

$=\frac{16}{5}·101$

Hence $m=101$