Question

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

Answer 1

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