If 1 3 + 2 3 + 3 3 . . . . . . upto  n   terms 1 · 3 + 2 · 5 + 3 · 7 + . . . .…

Given,

$\frac{1^3+2^3+3^3.......n^3}{1·3+2·5+3·7+....n\left(2n+1\right)}=\frac{9}{5}$

Now we know that $1^3+2^3+3^3.......n^3={\left[\frac{n\left(n+1\right)}{2}\right]}^2$

And $\Sigma n\left(2n+1\right)=2\Sigma n^2+\Sigma n$

$\Rightarrow \Sigma n\left(2n+1\right)=2\times \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}$

$\Rightarrow \Sigma n\left(2n+1\right)=\frac{n\left(n+1\right)\left(2n+1\right)}{3}+\frac{n\left(n+1\right)}{2}$

So putting the value in the given equation we get,

$\frac{1^3+2^3+3^3.......n^3}{1·3+2·5+3·7+....n\left(2n+1\right)}=\frac{9}{5}$

$\Rightarrow \frac{{\left[{\displaystyle \frac{n\left(n+1\right)}{2}}\right]}^2}{{\displaystyle \frac{{\displaystyle n\left(n+1\right)\left(2n+1\right)}}{{\displaystyle 3}}+\frac{{\displaystyle n\left(n+1\right)}}{{\displaystyle 2}}}}=\frac{9}{5}$

$\Rightarrow \frac{{\displaystyle \frac{1}{4}}\left[{\displaystyle n\left(n+1\right)}\right]}{{\displaystyle \frac{{\displaystyle \left(2n+1\right)}}{{\displaystyle 3}}+\frac{{\displaystyle 1}}{{\displaystyle 2}}}}=\frac{9}{5}$

$\Rightarrow \frac{{\displaystyle 6\left[n\left(n+1\right)\right]}}{{\displaystyle 4\left(2\left(2n+1\right)+3\right)}}=\frac{9}{5}$

$\Rightarrow \frac{{\displaystyle \left[n\left(n+1\right)\right]}}{{\displaystyle 2\left(2\left(2n+1\right)+3\right)}}=\frac{3}{5}$

$\Rightarrow 5\left[n\left(n+1\right)\right]=6\left(2\left(2n+1\right)+3\right)$

$\Rightarrow 5n^2+5n=6\left(4n+5\right)$

$\Rightarrow 5n^2+5n=24n+30$

$\Rightarrow 5n^2-19n-30=0$

$\Rightarrow 5n^2-25n+6n-30=0$

$\Rightarrow \left(5n+6\right)\left(n-5\right)=0$

$\Rightarrow n=5$