Mathematics - Binomial Theorem Question with Solution | TestHub

To find the coefficient of term independent of$x$in given expression, first let's find constant term and coefficient of$x^{-8}$in the expansion of$\mathrm{ }{\left(2x^2-\frac{3}{x^2}\right)}^6$
$T_{r+1}{= }^6{C}_r {\left(2x^2\right)}^{6-r}\times {\left(-\frac{3}{x^2}\right)}^r$
$T_{r+1}{= }^6{C}_r 2^{6-r}\times {\left(-3\right)}^r \times x^{12-4r}$
For constant term:$r=3$, hence constant term =${ }^6{C}_3 2^3\times {\left(-3\right)}^3\times x^0$
For coefficient of$x^{-8} : r=5$, hence coefficient of$x^{-8}{= }^6{C}_5\times 2^1{\left(-3\right)}^5\times x^{-8}$
Given expression can be written as
$\left(\frac{1}{60}-\frac{x^8}{81}\right)\left({ }^6{C}_3\times 2^3\times {\left(-3\right)}^3\times x^0+\dots + { }^6{C}_5\times 2^1\times {\left(-3\right)}^5\times x^{-8}+\dots \right)$
Hence, coefficient of term independent of$\mathrm{x}$
$=\frac{1}{60}{\times \mathrm{ }}^6{\mathrm{C}}_3\times 2^3{\left(-3\right)}^3-\frac{1}{81}{\times }^{\mathrm{ }\mathrm{ }6}{\mathrm{C}}_5\times 2^1\times {\left(-3\right)}^5$
$=\mathrm{ }-72+36=\mathrm{ }-36$