Mathematics - Binomial Theorem Question with Solution | TestHub

L.H.S.

$$=\sum _{r=0}^{n-1}{\left(\frac{n_{C_r}}{n_{C_r}+n_{C_{r+1}}}\right)}^3=\sum _{r=0}^{n-1}{\left(\frac{n_{C_r}}{n_{C_r}\left(1+\frac{n_{C_{r+1}}}{n_{C_r}}\right)}\right)}^3$$

$$=\sum _{r=0}^{n-1}{\left(\frac{r+1}{n+1}\right)}^3=\frac{n^2(n+1{)}^2}{4(n+1{)}^3}=\frac{n^2}{4(n+1)}$$

Now, we have $\frac{n^2}{4(n+1)}=\frac{4}{5}$.

This implies $5n^2-16n-16=0$.

Solving for $n$, we get $n=4$ or $n=-20$.

The answer is $n=4$.