Mathematics - Binomial Theorem Question with Solution | TestHub

Let $t_r=\frac{{}^n{C}_{r-1}}{{}^n{C}_r+{}^n{C}_{r-1}}$.

Using the identity ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$, we can rewrite $t_r$ as:

$$t_r=\frac{{}^n{C}_{r-1}}{{}^{n+1}{C}_r}$$

We also know that ${}^{n+1}{C}_r=\frac{n+1}{r}{}^n{C}_{r-1}$. Substituting this into the equation for $t_r$:

$$t_r=\frac{{}^n{C}_{r-1}}{\frac{n+1}{r}{}^n{C}_{r-1}}$$

Therefore, $t_r=\frac{r}{n+1}$.

Now, let $S={\sum }_{r=0}^n\{t_r{\}}^3$. Substituting the expression for $t_r$:

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

We can factor out $\frac{1}{(n+1{)}^3}$:

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

Using the formula for the sum of the first $n$ cubes, ${\sum }_{r=0}^n{r}^3={\left(\frac{n(n+1)}{2}\right)}^2$:

$$S=\frac{1}{(n+1{)}^3}{\left(\frac{n(n+1)}{2}\right)}^2$$

$$S=\frac{1}{(n+1{)}^3}\frac{n^2(n+1{)}^2}{4}$$

$$S=\frac{n^2}{4(n+1)}$$

We are given that $S=\frac{25}{24}$. This is only possible for $n=5$.