Question

Find the constant term in the expansion of ${(4x^2-3/2x)}^{12}$

Answer 1

${(4x^2-\frac{3}{2x})}^{12}$

For $(a+b{)}^n\to {}^n{C}_r{a}^{n-r}{b}^r$

$\to {}^{12}{C}_r\left(4{)}^{n-r}\right(x^2{)}^{n-r}{\left(\frac{-3}{2x}\right)}^r$

$(x^2{)}^{n-r}{\left(\frac{-1}{2x}\right)}^r$

$(-1{)}^r{x}^{2n-2r-r}$

$2n-3r=0$ $[\because n=12]$

$24=3r\Rightarrow r=8$

co-efficient will be ${}^{12}{C}_8(4{)}^{12-8}{\left(\frac{-3}{2}\right)}^8$

$\frac{12!}{8!4!}(4{)}^4\times \frac{3^8}{2^8}(-1{)}^8$

$=\frac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}\times 3^8$

$=495\times 3^8$ is constant term.