Prove that the coefficient of xn in the binomial expansion of

Solution:

To Prove : coefficient of $x^{n}$ in $(1+x)^{2 n}=2 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n-1}$

For $(1+x)^{2 n}$

$a=1, b=x$ and $m=2 n$

We have a formula,

$t_{r+1}=\left(\begin{array}{c}m \\ r\end{array}\right) a^{m-r} b^{r}$

$=\left(\begin{array}{c}2 \mathrm{n} \\ \mathrm{r}\end{array}\right)(1)^{2 \mathrm{n}-\mathrm{r}}(\mathrm{x})^{\mathrm{r}}$

$=\left(\begin{array}{c}2 \mathrm{n} \\ \mathrm{r}\end{array}\right)(\mathrm{x})^{\mathrm{r}}$

To get the coefficient of $x^{n}$, we must have,

$x^{n}=x^{r}$

- $r=n$

Therefore, the coefficient of $x^{n}=\left(\begin{array}{c}2 n \\ n\end{array}\right)$

$=\frac{(2 \mathrm{n}) !}{\mathrm{n} ! \times(2 \mathrm{n}-\mathrm{n}) !} \ldots\left(\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\frac{\mathrm{n} !}{\mathrm{r} ! \times(\mathrm{n}-\mathrm{r}) !}\right)$

$=\frac{(2 n) !}{n ! \times n !}$

$=\frac{2 \mathrm{n} \times(2 \mathrm{n}-1) !}{\mathrm{n} ! \times \mathrm{n}(\mathrm{n}-1) !} \ldots \ldots \ldots . .(\because \mathrm{n} !=\mathrm{n}(\mathrm{n}-1) !)$

$=\frac{2 \times(2 n-1) !}{n ! \times(n-1) !}$ ………cancelling n

Therefore, the coefficient of $x^{n}$ in $(1+x)^{2 n}$ $=\frac{2 \times(2 \mathrm{n}-1) !}{\mathrm{n} ! \times(\mathrm{n}-1) !}$ ………eq(1)

Now for $(1+x)^{2 n-1}$,

$a=1, b=x$ and $m=2 n-1$

We have formula,

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{m} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{m}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$=\left(\begin{array}{c}2 n-1 \\ r\end{array}\right)(1)^{2 n-1-r}(x)^{r}$

$=\left(\begin{array}{c}2 n-1 \\ r\end{array}\right)(x)^{r}$

To get the coefficient of $x^{n}$, we must have,

$X^{n}=X^{r}$

- $r=n$

Therefore, the coefficient of $x^{n}$ in $(1+x)^{2 n-1}=\left(\begin{array}{c}2 n-1 \\ n\end{array}\right)$

$=\frac{(2 n-1) !}{n ! \times(2 n-1-n) !}$

$=\frac{1}{2} \times \frac{2 \times(2 n-1) !}{n ! \times(n-1) !}$

…..multiplying and dividing by 2

Therefore

Coefficient of $x^{n}$ in $(1+x)^{2 n-1}=0 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n}$

Or coefficient of $x^{n}$ in $(1+x)^{2 n}=2 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n-1}$

Hence proved.