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Question Number 27809 by das47955@mail.com last updated on 15/Jan/18

$$\left(1\right)\mathbf{Find}\mathbf{the}\mathbf{term}\mathbf{independent}$$$$\mathbf{of}\mathbf{x}\mathbf{in}\mathbf{the}\mathbf{expansion}\mathbf{of}$$$${(\mathbf{x}-\frac{2}{\mathbf{x}})}^{10}$$

Answered by 8/mln(naing)060691 last updated on 15/Jan/18

$${(\mathrm{r}+1)}^{\mathrm{th}}\mathrm{term}{=}^{10}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{10-\mathrm{r}}{(-\frac{2}{\mathrm{x}})}^{\mathrm{r}}$$$${=}^{10}{\mathrm{C}}_{\mathrm{r}}.{\mathrm{x}}^{10-\mathrm{r}}{(-2)}^{\mathrm{r}}{\mathrm{x}}^{-\mathrm{r}}$$$${=}^{10}{\mathrm{C}}_{\mathrm{r}}{(-2)}^{\mathrm{r}}{\mathrm{x}}^{10-2\mathrm{r}}$$$$\mathrm{To}\mathrm{get}\mathrm{the}\mathrm{term}\mathrm{independent}\mathrm{of}\mathrm{x},$$$$\mathrm{put}10-2\mathrm{r}=0\Rightarrow \mathrm{r}=5$$$$\therefore \mathrm{the}\mathrm{term}\mathrm{independent}\mathrm{of}\mathrm{x}{=}^{10}{\mathrm{C}}_5{(-2)}^5$$

Commented by das47955@mail.com last updated on 15/Jan/18

$$$$$$\mathrm{wow}.\mathrm{i}\mathrm{need}\mathrm{this}\mathrm{process}$$

Commented by das47955@mail.com last updated on 15/Jan/18

$$$$$$\mathrm{really}\mathrm{thanks}.....$$

Answered by mrW2 last updated on 15/Jan/18

$${(x-\frac{2}{x})}^{10}=\underset{k=0}{\sum ^{10}}{C}_k^{10}{x}^k{(-\frac{2}{x})}^{10-k}$$$$=\underset{k=0}{\sum ^{10}}{C}_k^{10}{(-2)}^{10-k}{x}^{2k-10}$$$$2k-10=0$$$$k=5$$$$x-independenttermis$$$$C_5^{10}{(-2)}^5=-8064$$

Commented by das47955@mail.com last updated on 15/Jan/18

$$\mathbf{T}\mathrm{hanks}....$$

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