Question

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

Short answer

The coefficient of$x^k\text{is}\left(\begin{array}{c}100\\ -\frac{k}{2}+50\end{array}\right)\text{when}k\equiv 0(mod2)\text{and}-100\le k\le 100$

Step-by-step solution

Use Binomial theorem

Binomial theorem: binomial theorem, statement that for any positive integer, the nth power of the sum of two numbers x and y may be expressed as the sum of 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${\left(x+\frac{1}{x}\right)}^{100}$

$\begin{array}{r}n=100\\ y=\frac{1}{x}\end{array}$

Find the corresponding term

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

$\text{Let}k=100-2j\text{, then}-2j=k-100\text{or}$equivalently $j=-\frac{k}{2}+50$

Thus, the coefficient $x^k\text{is}\left(\begin{array}{c}100\\ j\end{array}\right)=\left(\begin{array}{c}100\\ -\frac{k}{2}+50\end{array}\right)$.