Question

Find the sixth term in the expansion ${\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)}^n$, if the binomial coefficient of the third term from the end is 45.

Answer 1

In the binomial expansion of ${\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)}^n$, there are (n + 1) terms.

The third term from the end in the expansion of ${\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)}^n$, is the third term from the beginning in the expansion of ${\left(x^{\frac{1}{3}}+y^{\frac{1}{2}}\right)}^n$.

∴ The binomial coefficient of the third term from the end = ${}^n{C}_2$

If is given that the binomial coefficient of the third term from the end is 45.

$$\begin{gathered} \therefore {}^n{C}_2=45 \\ \Rightarrow \frac{n\left(n-1\right)}{2}=45 \\ \Rightarrow n^2-n-90=0 \\ \Rightarrow \left(n-10\right)\left(n+9\right)=0 \\ \Rightarrow n=10\left(\because n\mathrm{cannot}\mathrm{be}\mathrm{negative}\right) \end{gathered}$$

Let T₆ be the sixth term in the binomial expansion of ${\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)}^n$. Then
$T_6{=}^n{C}_5{\left(y^{\frac{1}{2}}\right)}^{n-5}{\left(x^{\frac{1}{3}}\right)}^5{=}^{10}{C}_5{y}^{\frac{5}{2}}{x}^{\frac{5}{3}}=252y^{\frac{5}{2}}{x}^{\frac{5}{3}}$

Hence, the sixth term in the expansion of ${\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)}^n$, is $252y^{\frac{5}{2}}{x}^{\frac{5}{3}}$.