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Question

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


Solution

$$\left(x^2 + \dfrac{1}{x}\right)^{2n}$$
$$T_{r + 1} = ^{2n} Cr (x^2)^{2n - r} \left(\dfrac{1}{x}\right)^r$$
$$T_{r +1} = ^{2n} Cr \, x^{4n - 2r - r}$$
$$T_{r + 1} = ^{2n} Cr x^{4n - 3r}$$
$$4n - 3r = p$$
$$\Rightarrow 4n - p = 3r$$
$$\Rightarrow r = \dfrac{4n - p}{3}$$
$$\therefore$$ coefficient of $$x^p$$ is $$ ^{2n}C_{\dfrac{4n - p}{3}}$$
$$= \dfrac{(2n)!}{\left(\dfrac{4n - p}{3} \right) ! \left(\dfrac{2n + p}{3} \right)!}$$

Mathematics

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