Question

Find the $6^{\text{th}}$ term of the expansion $(y^{1/2}+x^{1/3}{)}^n,$ if the value of coefficient of $3^{\text{rd}}$ term from the end is $45$

• A. $232y^{5/2}{x}^{5/3}$
• B. $252y^{5/2}{x}^{5/3}$
• C. $232y^{5/3}{x}^{5/2}$
• D. $252y^{5/3}{x}^{5/2}$

Answer 1

The correct option is B $252y^{5/2}{x}^{5/3}$

Coefficient of $3^{\text{rd}}$ term from the beginning is always equal to the coefficient of $3^{\text{rd}}$term from the end.

Now,
$T_{r+1}={}^n{C}_r(y^{1/2}{)}^{(n-r)}\cdot (x^{1/3}{)}^r$
Coefficient of $T_3={}^n{C}_2=\frac{n(n-1)}{2}$

$\therefore \frac{n(n-1)}{2}=45\Rightarrow n^2-n-90=0\Rightarrow n^2-10n+9n-90=0\Rightarrow (n-10)(n+9)=0\Rightarrow n=10(\because n\in N)$

$\therefore T_6=T_{5+1}\Rightarrow T_6={}^{10}{C}_5(y^{1/2}{)}^{10-5}\cdot (x^{1/3}{)}^5\Rightarrow T_6=\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}\times y^{5/2}{x}^{5/3}\therefore T_6=252y^{5/2}{x}^{5/3}$