`
`
Question:
Find the sixth term of 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 .
Solution:
Given $\left(y^{1 / 2}+x^{1 / 3}\right)^{n}$
Also given that binomial coefficient of third term from the end $=45$ Therefore,
${ }^{n} C_{n-2}=45$
The above expression can be written as
${ }^{n} C_{2}=45$
$\Rightarrow \frac{n(n-1)(n-2) !}{2 !(n-2) !}=45$
$\Rightarrow \quad n(n-1)=90$
$\Rightarrow \quad n^{2}-n-90=0$
$\Rightarrow(n-10)(n+9)=0$
Therefore $n=10$
Now, sixth term $={ }^{10} C_{5}\left(y^{1 / 2}\right)^{10-5}\left(x^{1 / 3}\right)^{5}=252 y^{5 / 2} \cdot x^{5 / 3}$
.jpg)