Question

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

Answer 1

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

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

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

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

${\therefore }^n{C}_2=45$

$\Rightarrow \frac{n(n-1)}{2}=45$

$\Rightarrow n^2-n-90=0$

$\Rightarrow (n-10)(n+9)=0$

$\Rightarrow n=10(\because$ n cannot be negative)

let $T_6$ be the sixth term in the binomial expansion of ${(y^{\frac{1}{2}}+x^{\frac{1}{3}})}^n$. 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 $(y^{\frac{1}{2}}+x^{\frac{1}{3}})$, is $252y^{\frac{5}{2}}{x}^{\frac{5}{3}}$