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Question-140751

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Question Number 140751 by peter frank last updated on 12/May/21
Answered by Ar Brandon last updated on 12/May/21
(1+x)^n =Σ_(r=0) ^n ^n C_r x^r ^n C_(r−1) , ^n C_r , and ^n C_(r+1) are in AP ^n C_(r−1) + ^n C_(r+1) =2 ^n C_r ((n!)/((n−r+1)!(r−1)!))+((n!)/((n−r−1)!(r+1)!))=((2n!)/((n−r)!r!)) ((n!(r+1)r+n!(n−r+1)(n−r))/((n−r+1)!(r+1)!))=((2n!)/((n−r)!r!)) (((r+1)r+(n−r+1)(n−r))/((n−r+1)(r+1)))=2 r^2 +r+n^2 −nr−nr+r^2 +n−r=2(nr+n−r^2 −r+r+1) 4r^2 +n^2 −4nr−n−2=0 n^2 −n(4r+1)+4r^2 −2=0
$$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} =\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \\ $$$$\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}−\mathrm{1}} ,\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} ,\:\mathrm{and}\:\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}+\mathrm{1}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$$$\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}−\mathrm{1}} +\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}+\mathrm{1}} =\mathrm{2}\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} \\ $$$$\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)!\left(\mathrm{r}−\mathrm{1}\right)!}+\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{r}−\mathrm{1}\right)!\left(\mathrm{r}+\mathrm{1}\right)!}=\frac{\mathrm{2n}!}{\left(\mathrm{n}−\mathrm{r}\right)!\mathrm{r}!} \\ $$$$\frac{\mathrm{n}!\left(\mathrm{r}+\mathrm{1}\right)\mathrm{r}+\mathrm{n}!\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{r}\right)}{\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)!\left(\mathrm{r}+\mathrm{1}\right)!}=\frac{\mathrm{2n}!}{\left(\mathrm{n}−\mathrm{r}\right)!\mathrm{r}!} \\ $$$$\frac{\left(\mathrm{r}+\mathrm{1}\right)\mathrm{r}+\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{r}\right)}{\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)\left(\mathrm{r}+\mathrm{1}\right)}=\mathrm{2} \\ $$$$\mathrm{r}^{\mathrm{2}} +\mathrm{r}+\mathrm{n}^{\mathrm{2}} −\mathrm{nr}−\mathrm{nr}+\mathrm{r}^{\mathrm{2}} +\mathrm{n}−\mathrm{r}=\mathrm{2}\left(\mathrm{nr}+\mathrm{n}−\mathrm{r}^{\mathrm{2}} −\mathrm{r}+\mathrm{r}+\mathrm{1}\right) \\ $$$$\mathrm{4r}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} −\mathrm{4nr}−\mathrm{n}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{n}^{\mathrm{2}} −\mathrm{n}\left(\mathrm{4r}+\mathrm{1}\right)+\mathrm{4r}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$
Commented by peter frank last updated on 12/May/21
thank you
$${thank}\:{you} \\ $$

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