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 the value of $n$ is

• A. $10$
• B. $5$
• C. $20$
• D. $15$

Answer 1

The correct option is B $5$

${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$
$\Rightarrow \frac{{}^n{C}_{r-1}}{{}^n{C}_r+{}^n{C}_{r-1}}=\frac{{}^n{C}_{r-1}}{{}^{n+1}{C}_r}$
$=\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{(n+1)!}{r!(n+1-r)!}}$
$=\frac{r}{n+1}$

$\therefore \sum _{r=0}^n{\left(\frac{{}^n{C}_{r-1}}{{}^n{C}_r+{}^n{C}_{r-1}}\right)}^3=\sum _{r=0}^n{\left(\frac{r}{n+1}\right)}^3$
$=\frac{1}{(n+1{)}^3}\sum _{r=0}^n{r}^3$

$\because \sum _{r=1}^n{r}^3=\sum _{r=0}^n{r}^3={\left(\frac{n(n+1)}{2}\right)}^2$
$\therefore \frac{n^2}{4(n+1)}=\frac{25}{24}=\frac{5^2}{4(5+1)}$
$\Rightarrow n=5$