\CIE\0606\2020\w\paper 12 \no 5
Find the coefficient of $x^{2}$ in the expansion of $\left(x-\dfrac{3}{x}\right)\left(x+\dfrac{2}{x}\right)^{5}$.
\CIE\0606\2020\w\paper 12 \no 5
Find the coefficient of $x^{2}$ in the expansion of $\left(x-\dfrac{3}{x}\right)\left(x+\dfrac{2}{x}\right)^{5}$.
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$******$ Solution $*****$ $\def\iixi#1#2{\left(\begin{array}{c}#1\\#2\end{array}\right)}$ $\left(x+\dfrac{2}{x}\right)^5=x^5+\iixi 51 x^4\left(\dfrac{2}{x}\right)^1+\iixi{5}{2}x^3\left(\dfrac{2}{x}\right)^2+\iixi{5}{3}x^2\left(\dfrac{2}{x}\right)^3+\iixi{5}{4}x\left(\dfrac{2}{x}\right)^4+\left(\dfrac{2}{x}\right)^5$ $=x^5+10x^3+40x+\dfrac{80}{x}+\dfrac{80}{x^3}+\dfrac{32}{x^5}$ $\therefore\left(x-\dfrac{3}{x}\right)\left(x+\dfrac{2}{x}\right)^{5}=\left(x-\dfrac{3}{x}\right)\left(x^5+10x^3+40x+\dfrac{80}{x}+\dfrac{80}{x^3}+\dfrac{32}{x^5}\right)$ Coefficient of $x^2$ is $1\times 40+(-3)\times 10=10.$ $******$ End of Solution $*****$ |
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