91Ó°ÊÓ

Here ${(x-\frac{1}{x^2})}^{12}$is given

We have to find out the coefficient of x⁰ is 495

The Binomial theorem states that

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.......+\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \\ basedonthis{(x^2+\frac{1}{x})}^{12}becomes \\ {(x-\frac{1}{x^2})}^{12}=\left(\begin{array}{c}12\\ 0\end{array}\right){(-\frac{1}{x^2})}^0{\left(x\right)}^{12}+\left(\begin{array}{c}12\\ 1\end{array}\right)×{(-\frac{1}{x^2})}^1{×\left(x\right)}^{12-1}+\left(\begin{array}{c}12\\ 2\end{array}\right)×{(-\frac{1}{x^2})}^2{×\left(x\right)}^{12-2}+\left(\begin{array}{c}12\\ 3\end{array}\right)×{(-\frac{1}{x^2})}^3{×\left(x\right)}^{12-3}+\left(\begin{array}{c}12\\ 4\end{array}\right)×{(-\frac{1}{x^2})}^4{×\left(x\right)}^{12-4} \\ +\left(\begin{array}{c}12\\ 5\end{array}\right)×{(-\frac{1}{x^2})}^5{×\left(x\right)}^{12-5}+\left(\begin{array}{c}12\\ 6\end{array}\right)×{(-\frac{1}{x^2})}^6{×\left(x\right)}^{12-6}+\left(\begin{array}{c}12\\ 7\end{array}\right)×{(-\frac{1}{x^2})}^7{×\left(x\right)}^{12-7}+\left(\begin{array}{c}12\\ 8\end{array}\right)×{(-\frac{1}{x^2})}^8{×\left(x\right)}^{12-8} \\ +\left(\begin{array}{c}12\\ 9\end{array}\right)×{(-\frac{1}{x^2})}^9{×\left(x\right)}^{12-9}+\left(\begin{array}{c}12\\ 10\end{array}\right)×{(-\frac{1}{x^2})}^{10}{×\left(x\right)}^{12-10}+\left(\begin{array}{c}12\\ 11\end{array}\right)×{(-\frac{1}{x^2})}^{11}{×\left(x\right)}^{12-11}+\left(\begin{array}{c}12\\ 12\end{array}\right)×{(-\frac{1}{x^2})}^{12}{×\left(x\right)}^{12-12} \\ =\left(\begin{array}{c}12\\ 0\end{array}\right)x^{12}-\left(\begin{array}{c}12\\ 1\end{array}\right)x^9+\left(\begin{array}{c}12\\ 2\end{array}\right)x^6-\left(\begin{array}{c}12\\ 3\end{array}\right)x^3+\left(\begin{array}{c}12\\ 4\end{array}\right)x^0-\left(\begin{array}{c}12\\ 5\end{array}\right)x^{-3}+\left(\begin{array}{c}12\\ 6\end{array}\right)x^{-6}-\left(\begin{array}{c}12\\ 7\end{array}\right)x^{-9}+\left(\begin{array}{c}12\\ 8\end{array}\right)x^{-12}-\left(\begin{array}{c}12\\ 9\end{array}\right)x^{-15} \\ +\left(\begin{array}{c}12\\ 10\end{array}\right)x^{-18}-\left(\begin{array}{c}12\\ 11\end{array}\right)x^{-21}+\left(\begin{array}{c}12\\ 12\end{array}\right)x^{-24} \end{gathered}$$

The term containing x⁰ from the expansion is

$$\begin{gathered} \left(\begin{array}{c}12\\ 4\end{array}\right)x^{0,}hencecoefficientof{x}^{0}is\left(\begin{array}{c}12\\ 4\end{array}\right)=\frac{12!}{4!8!} \\ =\frac{12×11×10×9×8!}{4×3×2×1×8!} \\ =495 \end{gathered}$$