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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
1 year, 2 months ago

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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 \[\large ^{2n}C_{\frac{4n-p}{3}} = \dfrac{2n!}{\left(\dfrac{1}{3}(4n-p)\right)\large{!}\left(\dfrac{1}{3}(2n+p)\right)\large{!}}.\] Akhil Bansal · 1 year, 2 months ago

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