91Ó°ÊÓ

Here ${\left(x-\frac{2}{\sqrt{x}}\right)}^{10}$is given.

We have to find out thecoefficient of x⁴ in the expansion

By using the Binomial theorem we can expand the given expression.

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 \\ {(x-\frac{2}{\sqrt{x}})}^{10}=\left(\begin{array}{c}10\\ 0\end{array}\right){(-\frac{2}{\sqrt{x}})}^0{\left(x\right)}^{10}+\left(\begin{array}{c}10\\ 1\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^1{×\left(x\right)}^{10-1}+\left(\begin{array}{c}10\\ 2\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^2{×\left(x\right)}^{10-2}+\left(\begin{array}{c}10\\ 3\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^3{×\left(x\right)}^{10-3}+\left(\begin{array}{c}10\\ 4\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^4{×\left(x\right)}^{10-4} \\ +\left(\begin{array}{c}10\\ 5\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^5{×\left(x\right)}^{10-5}+\left(\begin{array}{c}10\\ 6\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^6{×\left(x\right)}^{10-6}+\left(\begin{array}{c}10\\ 7\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^7{×\left(x\right)}^{10-7}+\left(\begin{array}{c}10\\ 8\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^8{×\left(x\right)}^{10-8} \\ +\left(\begin{array}{c}10\\ 9\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^9{×\left(x\right)}^{10-9}+\left(\begin{array}{c}10\\ 10\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^{10}{×\left(x\right)}^{10-10} \\ =\left(\begin{array}{c}10\\ 0\end{array}\right)x^{10}-\left(\begin{array}{c}10\\ 1\end{array}\right)2x^{8.5}+\left(\begin{array}{c}10\\ 2\end{array}\right)4x^7-\left(\begin{array}{c}10\\ 3\end{array}\right)8x^{5.5}+\left(\begin{array}{c}10\\ 4\end{array}\right)16x^4-\left(\begin{array}{c}10\\ 5\end{array}\right)32x^{2.5}+\left(\begin{array}{c}10\\ 6\end{array}\right)64x^1-\left(\begin{array}{c}10\\ 7\end{array}\right)128x^{-1.5}+\left(\begin{array}{c}10\\ 8\end{array}\right)256x^{-3}-\left(\begin{array}{c}10\\ 9\end{array}\right)512x^{-4.5} \\ +\left(\begin{array}{c}10\\ 10\end{array}\right)1024x^{-6} \end{gathered}$$

$$\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 \\ then \\ {(x-\frac{2}{\sqrt{x}})}^{10}=\left(\begin{array}{c}10\\ 0\end{array}\right){(-\frac{2}{\sqrt{x}})}^0{\left(x\right)}^{10}+\left(\begin{array}{c}10\\ 1\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^1{×\left(x\right)}^{10-1}+\left(\begin{array}{c}10\\ 2\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^2{×\left(x\right)}^{10-2}+\left(\begin{array}{c}10\\ 3\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^3{×\left(x\right)}^{10-3}+\left(\begin{array}{c}10\\ 4\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^4{×\left(x\right)}^{10-4} \\ +\left(\begin{array}{c}10\\ 5\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^5{×\left(x\right)}^{10-5}+\left(\begin{array}{c}10\\ 6\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^6{×\left(x\right)}^{10-6}+\left(\begin{array}{c}10\\ 7\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^7{×\left(x\right)}^{10-7}+\left(\begin{array}{c}10\\ 8\end{array}\right)×{(-\frac{1}{\sqrt{x}})}^8{×\left(x\right)}^{10-8} \\ +\left(\begin{array}{c}10\\ 9\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^9{×\left(x\right)}^{10-9}+\left(\begin{array}{c}10\\ 10\end{array}\right)×{(-\frac{2}{\sqrt{x}})}^{10}{×\left(x\right)}^{10-10} \\ =\left(\begin{array}{c}10\\ 0\end{array}\right)x^{10}-\left(\begin{array}{c}10\\ 1\end{array}\right)2x^{8.5}+\left(\begin{array}{c}10\\ 2\end{array}\right)4x^7-\left(\begin{array}{c}10\\ 3\end{array}\right)8x^{5.5}+\left(\begin{array}{c}10\\ 4\end{array}\right)16x^4-\left(\begin{array}{c}10\\ 5\end{array}\right)32x^{2.5}+\left(\begin{array}{c}10\\ 6\end{array}\right)64x^{}-\left(\begin{array}{c}10\\ 7\end{array}\right)128x^{-1.5}+\left(\begin{array}{c}10\\ 8\end{array}\right)256x^{-3}-\left(\begin{array}{c}10\\ 9\end{array}\right)512x^{-4.5} \\ +\left(\begin{array}{c}10\\ 10\end{array}\right)1024x^{-6} \end{gathered}$$

From the given expansion, we get the coefficient x⁴ is

$$\begin{gathered} \left(\begin{array}{c}10\\ 4\end{array}\right)×16=\frac{10!}{6!4!}×16 \\ =\frac{10×9×8×7×6!}{4×3×2×1×6!}×16 \\ =3360 \end{gathered}$$