91Ó°ÊÓ

${(x^2+\frac{1}{x})}^{12}$is given.

We have to find out the term x⁰

By using the Binomial theorem we can expand

$$\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^2+\frac{1}{x})}^{12}=\left(\begin{array}{c}12\\ 0\end{array}\right){\left(\frac{1}{x}\right)}^0{\left(x2\right)}^{12}+\left(\begin{array}{c}12\\ 1\end{array}\right)×{\left(\frac{1}{x}\right)}^1{×\left(x^2\right)}^{12-1}+\left(\begin{array}{c}12\\ 2\end{array}\right)×{\left(\frac{1}{x}\right)}^2{×\left(x^2\right)}^{12-2}+\left(\begin{array}{c}12\\ 3\end{array}\right)×{\left(\frac{1}{x}\right)}^3{×\left(x^2\right)}^{12-3}+\left(\begin{array}{c}12\\ 4\end{array}\right)×{\left(\frac{1}{x}\right)}^4{×\left(x^2\right)}^{12-4} \\ +\left(\begin{array}{c}12\\ 5\end{array}\right)×{\left(\frac{1}{x}\right)}^5{×\left(x^2\right)}^{12-5}+\left(\begin{array}{c}12\\ 6\end{array}\right)×{\left(\frac{1}{x}\right)}^6{×\left(x^2\right)}^{12-6}+\left(\begin{array}{c}12\\ 7\end{array}\right)×{\left(\frac{1}{x}\right)}^7{×\left(x^2\right)}^{12-7}+\left(\begin{array}{c}12\\ 8\end{array}\right)×{\left(\frac{1}{x}\right)}^8{×\left(x^2\right)}^{12-8} \\ +\left(\begin{array}{c}12\\ 9\end{array}\right)×{\left(\frac{1}{x}\right)}^9{×\left(x^2\right)}^{12-9}+\left(\begin{array}{c}12\\ 10\end{array}\right)×{\left(\frac{1}{x}\right)}^{10}{×\left(x^2\right)}^{12-10}+\left(\begin{array}{c}12\\ 11\end{array}\right)×{\left(\frac{1}{x}\right)}^{11}{×\left(x^2\right)}^{12-11}+\left(\begin{array}{c}12\\ 12\end{array}\right)×{\left(\frac{1}{x}\right)}^{12}{×\left(x^2\right)}^{12-12} \\ =\left(\begin{array}{c}12\\ 0\end{array}\right)x^{24}+\left(\begin{array}{c}12\\ 1\end{array}\right)x^{21}+\left(\begin{array}{c}12\\ 2\end{array}\right)x^{18}+\left(\begin{array}{c}12\\ 3\end{array}\right)x^{15}+\left(\begin{array}{c}12\\ 4\end{array}\right)x^{12}+\left(\begin{array}{c}12\\ 5\end{array}\right)x^9+\left(\begin{array}{c}12\\ 6\end{array}\right)x^6+\left(\begin{array}{c}12\\ 7\end{array}\right)x^3+\left(\begin{array}{c}12\\ 8\end{array}\right)x^0+\left(\begin{array}{c}12\\ 9\end{array}\right)x^{-3} \\ +\left(\begin{array}{c}12\\ 10\end{array}\right)x^{-6}+\left(\begin{array}{c}12\\ 11\end{array}\right)x^{-9}+\left(\begin{array}{c}12\\ 12\end{array}\right)x^{-12} \end{gathered}$$

The term containing x⁰ from the expansion is$$\begin{gathered} \left(\begin{array}{c}12\\ 8\end{array}\right)x^0,thenthecoefficientof{x}^0=\left(\begin{array}{c}12\\ 8\end{array}\right) \\ =\frac{12!}{8!4!} \\ =\frac{12×11×10×9×8!}{4×3×2×1×8!} \\ =495 \end{gathered}$$