Name: Binomial Expansion: Illustration
Uploaded: 2019-07-19
Duration: 4 min 17 s
Description: We already know what Binomial Theorem is. Let's watch this video to observe an illustration that will help us understand Binomial Expansion better.

Question

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} .

Accepted Answer

Assume that $x^{n}$ occurs in the ${(r + 1)}^{t h}$ term of the expansion ${(1 + x)}^{2 n}$
So $T_{r + 1} = {}^{2 n}C_{r} {(1)}^{2 n - r} {(x)}^{r} = {}^{2 n}C_{r} x^{r}$
Comparing the terms of $x$ in $x^{n}$ and in $T_{r + 1}$ , we get $r = n$
So, the coefficient of $x^{n}$ in the expansion of ${(1 + x)}^{2 n}$ is ${}^{2 n}C_{n}$
So, ${}^{2 n}C_{n} = \frac{(2 n)!}{(2 n - n)! n!} = \frac{(2 n)!}{{(n!)}^{2}} \dots \dots \dots \dots (i)$

Assume that $x^{n}$ occurs in the ${(k + 1)}^{t h}$ term of the expansion ${(1 + x)}^{2 n - 1}$
So $T_{k + 1} = {}^{2 n - 1}C_{k} {(1)}^{2 n - 1 - k} {(x)}^{k}$ $= {}^{2 n - 1}C_{k} x^{k}$
Comparing the terms of $x$ in $x^{n}$ and in $T_{k + 1}$ , we get $k = n$
So, the coefficient of $x^{n}$ in the expansion of ${(1 + x)}^{2 n - 1}$ is ${}^{2 n - 1}C_{n}$
So, ${}^{2 n - 1}C_{n} = \frac{(2 n - 1)!}{(2 n - 1 - n)! n!}$
$= \frac{(2 n - 1)!}{(n - 1)! n!}$ $= \frac{2 n (2 n - 1)!}{2 n (n - 1)! n!}$ $= \frac{(2 n)!}{2 n! n!}$ $= \frac{(2 n)!}{2 {(n!)}^{2}} \dots \dots \dots (i i)$
From equation (i) and (ii)
$= \frac{1}{2} ({}^{2 n}C_{n}) = {}^{2 n - 1}C_{n}$
$\Rightarrow {}^{2 n}C_{n} = 2 \cdot {}^{2 n - 1}C_{n}$
Thus, 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.

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

Important Questions on Binomial Theorem

Learn and Prepare for any exam you want !

Prove that the coefficient of xn in the expansion of 1+x2n is twice the coefficient of xn in the expansion of 1+x2n-1.

Important Questions on Binomial Theorem

Learn and Prepare for any exam you want !

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