Find the coefficient of

Question:

Find the coefficient of $x^{n}$ in the expansion of $(1+x)(1-x)^{n}$.

 

Solution:

To find: the coefficient of $x^{n}$ in the expansion of $(1+x)(1-x)^{n}$.

Formula Used:

Binomial expansion of $(x+y)^{n}$ is given by,

$(\mathrm{x}+\mathrm{y})^{\mathrm{n}}=\sum_{r=0}^{n}\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}$

Thus,

$(1+x)(1-x)^{n}$

$=(1+x)\left(\left(\begin{array}{l}n \\ 0\end{array}\right)(-x)+\left(\begin{array}{l}n \\ 1\end{array}\right)(-x)^{1}\right.$

$\left.+\left(\begin{array}{l}n \\ 2\end{array}\right)(-x)^{2}+. .+\left(\begin{array}{c}n \\ n-1\end{array}\right)(-x)^{n-1}+\left(\begin{array}{l}n \\ n\end{array}\right)(-x)^{n}\right)$

Thus, the coefficient of $(\mathrm{x})^{\mathrm{n}}$ is,

${ }^{n} C_{n}-{ }^{n} C_{n-1}$ (If $n$ is even)

$-{ }^{n} C_{n}+{ }^{n} C_{n-1}$ (If $n$ is odd)

Thus, the coefficient of $(\mathrm{x})^{\mathrm{n}}$ is, ${ }^{n} \mathrm{C}_{n}-{ }^{n} \mathrm{C}_{n-1}$ (If $\mathrm{n}$ is even) and $-{ }^{n} \mathrm{C}_{n}+{ }^{n} \mathrm{C}_{n-1}$ (If $\mathrm{n}$ is odd)

 

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