Question

Find the coefficient of $x^4$ in the expansion of ${(2x^2+\frac{3}{x^3})}^7$

• A. ${}^7{C}_2{2}^5{3}^3$
• B. ${}^7{C}_2{2}^5{3}^2$
• C. ${}^7{C}_2{3}^5{2}^2$
• D. ${}^7{C}_3{2}^5{3}^2$

Answer 1

The correct option is B ${}^7{C}_2{2}^5{3}^2$

Given, ${(2x^2+\frac{3}{x^3})}^7$

To find coefficient of $x^4$

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^7{C}_r(2x^2{)}^{7-r}{\left(\frac{3}{x^3}\right)}^r$

$T_{r+1}={}^7{C}_r\left(2{)}^{7-r}\right(x{)}^{14-2r}\frac{(3{)}^r}{x^{3r}}$

$T_{r+1}={}^7{C}_r\left(2{)}^{7-r}\right(3{)}^r\times (x{)}^{14-5r}$ ........$\left(1\right)$

given to find coefficient of $x^4$

$14-5r=4$

$14-4=5r$

$5r=10$

$r=2$

so putting $r=2$ in $\left(1\right)$

$T_{r+1}={}^7{C}_2\left(2{)}^{7-2}\right(3{)}^2(x{)}^4$

$T_{r+1}={}^7{C}_2{2}^5{3}^2{x}^4$

$\therefore$ So, coefficient of $x^4$ in the expansion of ${(2x^2+\frac{3}{x^3})}^7$ is ${}^7{C}_2{2}^5{3}^2$.