Question

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

• A. $252y^{5/2}\cdot x^{5/3}$
• B. $262y^{2/5}\cdot x^{5/3}$
• C. $250y^{5/2}\cdot x^{5/3}$
• D. $252y^{3/2}\cdot x^{5/3}$

Answer 1

Answer: (A)

Given expansion is $(y^{1/2}+x^{1/3}{)}^n$
Since $T_6=T_{5+1}={}^n{C}_5(y^{1/2}{)}^{n-5}(x^{1/3}{)}^5...(i)$
Since binomial coefficient of third term from the end $=$
Binomial coefficient of third term from the beginning $={}^n{C}_2$
$\therefore {}^n{C}_2=45$
$\Rightarrow \frac{n\left(n-1\right)\left(n-2\right)!}{2!\left(n-2\right)!}=45$
$\Rightarrow n\left(n-1\right)=90$
$\Rightarrow n^2-n-90=0$
$\Rightarrow n=10$
$\therefore$ From $\left(i\right),T_6={}^{10}{C}_5{y}^{5/2}{x}^{5/3}$
$=252y^{5/2}\cdot x^{5/3}$