Question

The coefficient of $x^4$ in the expansion of ${(4x^2+\frac{3}{x})}^8$ is

• A. ${}^8{C}_5{12}^5$
• B. ${}^8{C}_4{12}^4$
• C. ${}^8{C}_3{12}^3$
• D. ${}^8{C}_6{12}^6$

Answer 1

The correct option is B ${}^8{C}_4{12}^4$

Given,

${(4x^2+\frac{3}{x})}^8$

General term,

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

$T_{r+1}={}^8{C}_r(4x^2{)}^{8-r}{\left(\frac{3}{x}\right)}^r$

$T_{r+1}={}^8{C}_r\left(4{)}^{8-r}\right(x{)}^{16-2r}\frac{3^r}{x^r}$

$T_{r+1}={}^8{C}_r\left(4{)}^{8-r}\right(3{)}^r(x{)}^{16-3r}$

To find coefficient of $x^4$

$16-3r=4$

$12=3r$

$r=4$

$T_5={}^8{C}_4(4{)}^{8-4}{3}^4$

$T_5={}^8{C}_4(4{)}^4\times 3^4$

$T_5={}^8{C}_4(12{)}^4$

Hence correct option is B.