Question

Give a formula for the coefficient of${\mathit{x}}^{\mathbf{k}}$in the expansion of${\left(x^2-1/x\right)}^{\mathbf{100}}$, where Kis an integer.

Short answer

The coefficient of$x^k\text{is}\left(\begin{array}{c}100\\ (200-k)/3\end{array}\right)(-1{)}^{(200-k)/3}\text{when}k\equiv 2(mod3)\text{and}-100\le k\le 200$

Step-by-step solution

Use Binomial theorem

Binomial theorem: binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers x and y may be expressed as the sum of n + 1 terms of the form.

$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\left(\begin{array}{l}n\\ j\end{array}\right){\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{\mathit{y}}^{\mathbf{j}}$

The term is${\left(x^2-\frac{1}{x}\right)}^{100}$

$n=100$

Find the corresponding term

$\begin{array}{r}{\left(x^2-\frac{1}{x}\right)}^{100}=\sum _{j=0}^{100} \left(\begin{array}{c}100\\ j\end{array}\right){\left(x^2\right)}^{100-j}{\left(-\frac{1}{x}\right)}^j\\ =\sum _{j=0}^{100} \left(\begin{array}{c}100\\ j\end{array}\right)x^{2(100-j)}(-1{)}^j{x}^{-j}\\ =\sum _{j=0}^{100} \left(\begin{array}{c}100\\ j\end{array}\right)(-1{)}^j{x}^{200-2j-j}\\ =\sum _{j=0}^{100} \left(\begin{array}{c}100\\ j\end{array}\right)(-1{)}^j{x}^{200-3j}\end{array}$

Let , $k=200-3j\text{, then}-3j=k-200\text{or equivalently}j=\frac{200-k}{3}\text{.}$

Thus, the coefficient of $x^k\text{is}\left(\begin{array}{c}100\\ j\end{array}\right)(-1{)}^j=\left(\begin{array}{c}100\\ (200-k)/3\end{array}\right)(-1{)}^{(200-k)/3}\text{.}$