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Question Number 60406 by Tawa1 last updated on 20/May/19

$$\mathrm{n}\in {\mathbb{Z}}^{+},\mathrm{Find}\mathrm{the}\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^{-1}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}{(1+\mathrm{x})}^{\mathrm{n}}{(1+\frac{1}{\mathrm{x}})}^{\mathrm{n}}$$

Answered by mr W last updated on 20/May/19

$${(1+x)}^n{(1+\frac{1}{x})}^n$$$$=\frac{1}{x^n}{(1+x)}^{2n}$$$$=\frac{1}{x^n}\underset{k=0}{\sum ^{2n}}{C}_k^{2n}{x}^k$$$$=\underset{k=0}{\sum ^{2n}}{C}_k^{2n}{x}^{k-n}$$$$fork-n=-1\Rightarrow k=n-1$$$$coef.of{x}^{-1}is{C}_{n-1}^{2n}$$

Commented by Tawa1 last updated on 20/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{but}\mathrm{the}\mathrm{sum}\mathrm{is}\mathrm{confusing}$$

Commented by Tawa1 last updated on 20/May/19

$$\mathrm{I}\mathrm{now}\mathrm{understand}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 20/May/19

$$\mathrm{But}\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{sum}\mathrm{this}.$$$$\stackrel{\mathrm{n}}{}{\mathrm{C}}_0\stackrel{\mathrm{n}}{}{\mathrm{C}}_1+\stackrel{\mathrm{n}}{}{\mathrm{C}}_1\stackrel{\mathrm{n}}{}{\mathrm{C}}_2+\stackrel{\mathrm{n}}{}{\mathrm{C}}_2\stackrel{\mathrm{n}}{}{\mathrm{C}}_3+...+\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{r}}\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{r}+1}$$

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