Prove that the coefficient of $x n$ in the binomial expansion of $(1+x)^{2 n}$ is twice the coefficient of $x^{n}$ in the binomial expansion of $(1+x)^{2 n-1}$.
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.
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