Question

If ${\sum }_{r=0}^{n-1}{\left(\frac{{{}^{n}C}_r}{{{}^{n}C}_r+{{}^{n}C}_{r+1}}\right)}^3=\frac{4}{5}$ then $n=$

• A. $4$
• B. $6$
• C. $8$
• D. None of these

Answer 1

Answer: (A)

$4$

We know that $\left(\begin{array}{c}n\\ r+1\end{array}\right)+\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n+1\\ r+1\end{array}\right)$,

$⟹{\sum }_0^{n-1}{\left(\frac{\left(\begin{array}{c}n\\ r\end{array}\right)}{\left(\begin{array}{c}n+1\\ r+1\end{array}\right)}\right)}^3={\sum }_0^{n-1}{\left(\frac{r+1}{n+1}\right)}^3$

$⟹\frac{1^3+2^3+3^3+...+n^3}{{(n+1)}^3}⟹\frac{n^2{(n+1)}^2}{4{(n+1)}^3}⟹\frac{n^2}{4(n+1)}$

Given that $\frac{n^2}{4(n+1)}=\frac{4}{5}$

$⟹5n^2-16n-16=0⟹(n-4)(5n+4)=0$

As $n$ is a natural number

$⟹n=4$