Binomial Theorem

If $x^p$ occurs in the expansion of $\left(x^{2}+\dfrac{1}{x}\right)^{2n}$ ,prove that its coefficient is $\dfrac{2n!}{\left(\dfrac{1}{3}(4n-p)\right)\large{!}\left(\dfrac{1}{3}(2n+p)\right)\large{!}}.$

#Algebra

Easy Math Editor

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Comments

Let $T_{r+1}$ denotes the general term of given expression.
$\large T_{r+1} = ^{2n}C_r x^{2(2n-r)}\left(\dfrac{1}{x}\right)^r$ $\large T_{r+1} = ^{2n}C_r x^{4n -3r}$

For coefficient of $x^{p}$
$\large p = 4n -3r \Rightarrow r = \dfrac{4n - p}{3}$ Therefore, Coefficient of $x^p$ is $\large ^{2n}C_{\frac{4n-p}{3}} = \dfrac{2n!}{\left(\dfrac{1}{3}(4n-p)\right)\large{!}\left(\dfrac{1}{3}(2n+p)\right)\large{!}}.$

Akhil Bansal - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$