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

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Solution

It is known that $(r + 1)^{t h}$ term, $(T_{r + 1})$ , in the binomial expansion of $(a + b)^{n}$ is given by

$T_{r + 1} =^{n} C_{1} a^{n - r} b^{r}$

Assuming that $x^{n}$ occurs in the $(r + 1)^{t h}$ term of the expansion pf $(1 + x)^{2 n}$ , we obtain

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

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 $(k + 1)^{t h}$ term of the expansion of $(1 + x)^{2 n - 1}$ , we obtain

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

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

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

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

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

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

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

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

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

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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 $(1 + x)^{2 n - 1}$ .

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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

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