# Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .

Question:

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

Solution:

It is known that $(r+1)^{\text {th }}$ term $_{,}\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1}={ }^{n} C_{r} a^{n-t} b^{t}$.

Assuming that $x^{n}$ occurs in the $(r+1)^{\text {th }}$ term of the expansion of $(1+x)^{2 n}$, we obtain

$\mathrm{T}_{\mathrm{r}+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}(1)^{2 \mathrm{n}-\mathrm{t}}(\mathrm{x})^{\mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\mathrm{r}}$

Comparing the indices of $x$ in $x^{n}$ and in $T_{r+1}$, we obtain

$r=n$

Therefore, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ is

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

Assuming that $x^{n}$ occurs in the $(k+1)^{\text {th }}$ term of the expansion $(1+x)^{2 n-1}$, we obtain

$T_{k+1}={ }^{2 n-1} C_{k}(1)^{2 n-1-k}(x)^{k}={ }^{2 n-1} C_{k}(x)^{k}$

Comparing the indices of $x$ in $x^{n}$ and $T_{k+1}$, we obtain

$k=n$

Therefore, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n-1}$ is

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

$=\frac{2 n \cdot(2 n-1) !}{2 n \cdot n !(n-1) !}=\frac{(2 n) !}{2 \cdot n ! n !}=\frac{1}{2}\left[\frac{(2 n) !}{(n !)^{2}}\right]$

From (1) and (2), it is observed that

$\frac{1}{2}\left({ }^{2 n} C_{n}\right)={ }^{2 n-1} C_{n}$

$\Rightarrow^{2 n} C_{n}=2\left({ }^{2 n-1} C_{n}\right)$

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

Hence, proved.