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Question Number 31494 by abdo imad last updated on 09/Mar/18

$$provethat{x}^{2}divide{(x+1)}^{\underset{}{n}}-nx-1.nintegr.$$

Commented by Rasheed.Sindhi last updated on 10/Mar/18

$${(\mathrm{x}+1)}^{\mathrm{n}}-\mathrm{nx}-1$$$${\mathrm{x}}^{\mathrm{n}}+{\mathrm{nx}}^{\mathrm{n}-1}+...\left(\begin{array}{c}\mathrm{n}\\ \mathrm{n}-2\end{array}\right){\mathrm{x}}^2+\mathrm{nx}+1-\mathrm{nx}-1$$$${\mathrm{x}}^{\mathrm{n}}+{\mathrm{nx}}^{\mathrm{n}-1}+...\left(\begin{array}{c}\mathrm{n}\\ \mathrm{n}-2\end{array}\right){\mathrm{x}}^2$$$${\mathrm{x}}^2({\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{nx}}^{\mathrm{n}-3}+...\left(\begin{array}{c}\mathrm{n}\\ \mathrm{n}-2\end{array}\right))$$$$\mathrm{Hence}{\mathrm{x}}^2\mathrm{divides}{(\mathrm{x}+1)}^{\mathrm{n}}-\mathrm{nx}-1$$

Commented by abdo imad last updated on 09/Mar/18

$$letrememberthistheoremarootsofp\left(x\right)withordrm\iff$$$$p\left(a\right)=p^{{}^{\prime }}\left(a\right)=...=p^{m-1}\left(a\right)=0and{p}^{\left(m\right)}\left(a\right)\ne 0andif$$$$p\left(a\right)=p^{{}^{\prime }}\left(a\right)=...=p^{(m-1)}\left(a\right)=0\Rightarrow {(x-a)}^{m}dividep\left(x\right)$$$$herewehavep\left(x\right)={(x+1)}^n-nx-1\Rightarrow p\left(0\right)=0$$$$p^{{}^{\prime }}\left(x\right)=n{(x+1)}^{n-1}-n\Rightarrow p^{{}^{\prime }}\left(0\right)=n-n=0\Rightarrow x^{2}dividep\left(x\right)$$$$plusthatwehave{p}^{{}^{″}}\left(x\right)=n(n-1){(x+1)}^{n-2}\Rightarrow p^{{}^{″}}\left(p0\right)=n(n-1)$$$$p^{{}^{″}}\left(0\right)\ne 0\Rightarrow 0isrootofp\left(x\right)arordre2.$$$$$$

Commented by abdo imad last updated on 09/Mar/18

$$youranswerisalsocorrectsirrachid...$$

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