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Question Number 141024 by mathocean1 last updated on 15/May/21

$$GivenX={(\sqrt{\sqrt{x}}+\frac{1}{\sqrt{\sqrt{x}}})}^n.$$$$Whatisthecoefficientof{x}^{5-\frac{\mathrm{n}}{4}?}$$$$\mathrm{propositions}:$$$$a.5\frac{n!}{10!}$$$$b.\frac{n!}{5!}$$$$c.\left(\begin{array}{c}n\\ 10\end{array}\right)$$$$d.\left(\begin{array}{c}n\\ 5\end{array}\right)$$

Answered by Ar Brandon last updated on 15/May/21

$$X={(\sqrt{\sqrt{\mathrm{x}}}+\frac{1}{\sqrt{\sqrt{\mathrm{x}}}})}^{\mathrm{n}},\mathrm{let}\mathrm{x}={\mathrm{u}}^4.\mathrm{Find}\mathrm{coef}\left({\mathrm{u}}^{20-\mathrm{n}}\right)$$$$={(\mathrm{u}+\frac{1}{\mathrm{u}})}^{\mathrm{n}}=\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{r}}{\mathrm{u}}^{\mathrm{r}}{\left(\frac{1}{\mathrm{u}}\right)}^{\mathrm{n}-\mathrm{r}}=\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{r}}{\mathrm{u}}^{2\mathrm{r}-\mathrm{n}}$$$$\mathrm{r}=10\Rightarrow \mathrm{coef}\left({\mathrm{u}}^{20-\mathrm{n}}\right)=\stackrel{\mathrm{n}}{}{\mathrm{C}}_{10}=\left(\begin{array}{c}\mathrm{n}\\ 10\end{array}\right)$$

Answered by iloveisrael last updated on 15/May/21

$$X=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right){\left(x^{\frac{1}{4}}\right)}^k{\left(x^{-\frac{1}{4}}\right)}^{n-k}$$$$\Rightarrow x^{\frac{k+k-n}{4}}=x^{5-\frac{n}{4}};\frac{2k-n}{4}=\frac{20-n}{4}$$$$k=10\Rightarrow \mathrm{coefficient}=\left(\begin{array}{c}n\\ 10\end{array}\right)$$

Answered by mathocean1 last updated on 15/May/21

$$thankyousirs$$

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