# Binomial Theorem

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

Note by Rohit Udaiwal
2 years, 6 months ago

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## Comments

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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 $\large ^{2n}C_{\frac{4n-p}{3}} = \dfrac{2n!}{\left(\dfrac{1}{3}(4n-p)\right)\large{!}\left(\dfrac{1}{3}(2n+p)\right)\large{!}}.$

- 2 years, 6 months ago

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