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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