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# Find the sixth term of the expansion

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 $={ }^{10} C_{5}\left(y^{1 / 2}\right)^{10-5}\left(x^{1 / 3}\right)^{5}=252 y^{5 / 2} \cdot x^{5 / 3}$