Question

If $\frac{1^3+2^3+3^3+\dots \text{ up to }n\text{ terms }}{1\cdot 3+2\cdot 5+3\cdot 7+\dots \text{ up to }n\text{ terms }}=\frac{9}{5}$, then the value of $n$ is

Answer 1

Answer: 5

$1^3+2^3+3^3\dots .\cdot +n^3={\left(\frac{n(n+1)}{2}\right)}^2$
$1\cdot 3+2\cdot 5+3\cdot 7+\dots \dots +n\text{ terms }=$
$\sum _{r=1}^{n}r(2r+1)=\sum _{r=1}^n\left(2r^2+r\right)$
$=\frac{2\cdot n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{n(n+1)}{6}(2(2n+1)+3)$
$=\frac{n(n+1)}{2}\times \frac{(4n+5)}{3}$
$=\frac{\frac{n^2(n+1{)}^2}{4}}{\frac{n(n+1)}{2}\times \frac{(4n+5)}{3}}=\frac{9}{5}$
$\Rightarrow \frac{5n(n+1)}{2}=\frac{9(4n+5)}{3}$
$\Rightarrow 15n(n+1)=18(4n+5)$
$\Rightarrow 15n^2+15n=72n+90$
$\Rightarrow 15n^2-57n-90=0\Rightarrow 5n^2-19n-30=0$
$\Rightarrow (n-5)(5n+6)=0$
$\Rightarrow n=\frac{-6}{5}\text{ or }5$
$\Rightarrow n=5\text{. }$