Question Number 141024 by mathocean1 last updated on 15/May/21
$${Given}\:{X}=\left(\sqrt{\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{{x}}}}\right)^{{n}} . \\ $$$${What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}−\frac{\mathrm{n}}{\mathrm{4}}\:?} \\ $$$$\mathrm{propositions}: \\ $$$${a}.\:\:\:\:\mathrm{5}\frac{{n}!}{\mathrm{10}!} \\ $$$${b}.\:\:\:\:\frac{{n}!}{\mathrm{5}!} \\ $$$${c}.\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$$${d}.\:\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix} \\ $$
Answered by Ar Brandon last updated on 15/May/21
$${X}=\left(\sqrt{\sqrt{\mathrm{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{x}}}}\right)^{\mathrm{n}} ,\:\mathrm{let}\:\mathrm{x}=\mathrm{u}^{\mathrm{4}} .\:\mathrm{Find}\:\mathrm{coef}\left(\mathrm{u}^{\mathrm{20}−\mathrm{n}} \right) \\ $$$$\:\:\:\:=\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{u}}\right)^{\mathrm{n}} =\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} \mathrm{u}^{\mathrm{r}} \left(\frac{\mathrm{1}}{\mathrm{u}}\right)^{\mathrm{n}−\mathrm{r}} =\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} \mathrm{u}^{\mathrm{2r}−\mathrm{n}} \\ $$$$\mathrm{r}=\mathrm{10}\:\Rightarrow\mathrm{coef}\left(\mathrm{u}^{\mathrm{20}−\mathrm{n}} \right)=\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{10}} =\begin{pmatrix}{\mathrm{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$
Answered by iloveisrael last updated on 15/May/21
$${X}=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)^{{k}} \left({x}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)^{{n}−{k}} \\ $$$$\Rightarrow{x}^{\frac{{k}+{k}−{n}}{\mathrm{4}}} =\:{x}^{\mathrm{5}−\frac{{n}}{\mathrm{4}}} \:;\:\frac{\mathrm{2}{k}−{n}}{\mathrm{4}}\:=\:\frac{\mathrm{20}−{n}}{\mathrm{4}} \\ $$$${k}=\mathrm{10}\:\Rightarrow\:\mathrm{coefficient}\:=\:\begin{pmatrix}{\:{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$
Answered by mathocean1 last updated on 15/May/21
$${thank}\:{you}\:{sirs} \\ $$