Question

If ${\sum }_{r=0}^n{\left\{\frac{{{}^{n}C}_{r-1}}{{{}^{n}C}_r+{{}^{n}C}_{r-1}}\right\}}^3=\frac{25}{24}$, then $n$ is equal to ______

Answer 1

$\to As{}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$

$\Rightarrow \sum _{r=0}^n{\left\{\frac{{}^n{C}_{r-1}}{{}^{n+1}{C}_r}\right\}}^3=\frac{25}{24}$

$\Rightarrow \sum _{r=0}^n\{\frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n+1-r)!}{(n+1)!}\}=\frac{25}{24}$

$\Rightarrow \sum _{r=0}^n(\frac{r}{n+1}{)}^3=\frac{25}{24}$

$\Rightarrow \frac{\sum r^3}{(n+1{)}^3}=\frac{25}{24}$

$\Rightarrow {\left[\frac{n(n+1)}{2}\right]}^2\times \frac{1}{(n+1{)}^3}=\frac{25}{24}$

$\Rightarrow 24n^2=2\times 2(n+1)25$

$\Rightarrow 6n^2=25n+25$

$\Rightarrow 6n^2-25n-25=0$

$\Rightarrow n=\frac{25\pm \sqrt{625-4\left(6\right)\left(25\right)}}{12}$

$\Rightarrow n=\frac{25\pm 5\sqrt{25-24}}{12}$

$\Rightarrow n=\frac{25+5}{12}$ or $\frac{25-5}{12}$