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Question Number 121890 by oustmuchiya@gmail.com last updated on 12/Nov/20

Commented by liberty last updated on 12/Nov/20

$$\left(\mathrm{v}\right){({\mathrm{x}}^2+{\mathrm{x}}^{-1})}^{12}=\underset{\mathrm{r}=0}{\sum ^{12}}\left(\begin{array}{c}12\\ \mathrm{r}\end{array}\right){\left({\mathrm{x}}^2\right)}^{12-\mathrm{r}}.{\left({\mathrm{x}}^{-1}\right)}^{\mathrm{r}}$$$$\mathrm{the}\mathrm{term}\mathrm{independent}\mathrm{of}\mathrm{x},\mathrm{we}\mathrm{get}\mathrm{from}$$$$\Rightarrow {\mathrm{x}}^{24-2\mathrm{r}}.{\mathrm{x}}^{-\mathrm{r}}={\mathrm{x}}^0;24-3\mathrm{r}=0,\mathrm{r}=8$$$$\mathrm{gives}\left(\begin{array}{c}12\\ 8\end{array}\right){\left({\mathrm{x}}^2\right)}^4{\left({\mathrm{x}}^{-1}\right)}^8=\frac{8.7.6.5}{4.3.2.1}=70.▴$$

Answered by bemath last updated on 12/Nov/20

$$\left(iv\right){(1+x)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)1^{n-k}.x^k$$$$thethirdtermwhenk=2$$$$sothecooefficientofthethirdterm$$$$equalto\left(\begin{array}{c}n\\ 2\end{array}\right)=\frac{n(n-1)}{2}=21$$$$\Rightarrow n^2-n-42=0$$$$\Rightarrow (n-7)(n+6)=0\to n=7.⧫$$

Answered by TANMAY PANACEA last updated on 12/Nov/20

$$i){(1-\frac{x}{2})}^5$$$$=1-5C_1\left(\frac{x}{2}\right)+5C_2{\left(\frac{x}{2}\right)}^2-5C_3{\left(\frac{x}{2}\right)}^3+5C_4{\left(\frac{x}{2}\right)}^4-{\left(\frac{x}{2}\right)}^5$$$$ii){(1+\frac{x}{2})}^5+{(1-\frac{x}{2})}^5\Lleftarrow noteoddtermscancelled$$$$=2[1+5C_2{\left(\frac{x}{2}\right)}^2+5C_4{\left(\frac{x}{2}\right)}^4]$$$$iii)putx=0.1tofind{(1.05)}^5+{(0.95)}^4$$$$$$$$$$$$$$

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