First slide

Binomial theorem for positive integral Index

Question

$The coefficient of x^{n} in {(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!})}^{2}$

Moderate

A

$\frac{2^{n}}{n!}$
B

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

$n!$
D

$\frac{2^{n}}{(n + 1)!}$

Solution

${(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!})}^{2}$
$= (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!}) (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots + \frac{x^{n - 1}}{(n - 1)!} + \frac{x^{n}}{n!})$
$coefficient of x^{n}$
$\begin{array}{l} = 1 \cdot \frac{1}{n!} + \frac{1}{1!} \cdot \frac{1}{(n - 1)!} + \frac{1}{2!} \cdot \frac{1}{(n - 2)!} + \dots + \frac{1}{n! 1} \\ = \frac{1}{n!} (\frac{n!}{n!} + \frac{n!}{(n - 1)! 1!} + \frac{n!}{(n - 2)! 2!} + \dots + \frac{n!}{n!}) \\ = \frac{1}{n!} (^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n}) \\ = \frac{2^{n}}{n!} \end{array}$

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