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Question Number 31494 by abdo imad last updated on 09/Mar/18
prove that x^2 divide (x+1)^n_ −nx−1 .nintegr.
$${prove}\:{that}\:{x}^{\mathrm{2}} \:{divide}\:\left({x}+\mathrm{1}\right)^{\underset{} {{n}}} \:−{nx}−\mathrm{1}\:.{nintegr}. \\ $$
Commented by Rasheed.Sindhi last updated on 10/Mar/18
(x+1)^n −nx−1 x^n +nx^(n−1) +... ((( n)),((n−2)) )x^2 +nx+1−nx−1 x^n +nx^(n−1) +... ((( n)),((n−2)) )x^2 x^2 (x^(n−2) +nx^(n−3) +... ((( n)),((n−2)) ) ) Hence x^2 divides (x+1)^n −nx−1
$$\:\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} −\mathrm{nx}−\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{n}} +\mathrm{nx}^{\mathrm{n}−\mathrm{1}} +…\begin{pmatrix}{\:\:\:\mathrm{n}}\\{\mathrm{n}−\mathrm{2}}\end{pmatrix}\mathrm{x}^{\mathrm{2}} +\mathrm{nx}+\mathrm{1}−\mathrm{nx}−\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{n}} +\mathrm{nx}^{\mathrm{n}−\mathrm{1}} +…\begin{pmatrix}{\:\:\:\mathrm{n}}\\{\mathrm{n}−\mathrm{2}}\end{pmatrix}\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{nx}^{\mathrm{n}−\mathrm{3}} +…\begin{pmatrix}{\:\:\:\mathrm{n}}\\{\mathrm{n}−\mathrm{2}}\end{pmatrix}\:\right) \\ $$$$\mathrm{Hence}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{divides}\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} −\mathrm{nx}−\mathrm{1} \\ $$
Commented by abdo imad last updated on 09/Mar/18
let remember this theorem a roots of p(x) with ordr m⇔ p(a)=p^′ (a)=...=p^(m−1) (a)=0 and p^((m)) (a)≠0 and if p(a)=p^′ (a)=...=p^((m−1)) (a)=0 ⇒(x−a)^m divide p(x) here we have p(x)=(x+1)^n −nx −1 ⇒p(0)=0 p^′ (x)=n(x+1)^(n−1) −n ⇒p^′ (0)=n−n=0 ⇒x^2 divide p(x) plus that we have p^(′′) (x)=n(n−1)(x+1)^(n−2) ⇒p^(′′) (p0)=n(n−1) p^(′′) (0)≠0 ⇒0 is root of p(x) ar ordre 2.
$${let}\:{remember}\:{this}\:{theorem}\:\:{a}\:{roots}\:{of}\:{p}\left({x}\right)\:{with}\:{ordr}\:{m}\Leftrightarrow \\ $$$${p}\left({a}\right)={p}^{'} \left({a}\right)=…={p}^{{m}−\mathrm{1}} \left({a}\right)=\mathrm{0}\:{and}\:{p}^{\left({m}\right)} \left({a}\right)\neq\mathrm{0}\:\:{and}\:{if} \\ $$$${p}\left({a}\right)={p}^{'} \left({a}\right)=…={p}^{\left({m}−\mathrm{1}\right)} \left({a}\right)=\mathrm{0}\:\Rightarrow\left({x}−{a}\right)^{{m}} \:{divide}\:{p}\left({x}\right) \\ $$$${here}\:{we}\:{have}\:{p}\left({x}\right)=\left({x}+\mathrm{1}\right)^{{n}} \:−{nx}\:−\mathrm{1}\:\Rightarrow{p}\left(\mathrm{0}\right)=\mathrm{0}\: \\ $$$${p}^{'} \left({x}\right)={n}\left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}} \:−{n}\:\Rightarrow{p}^{'} \left(\mathrm{0}\right)={n}−{n}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} {divide}\:{p}\left({x}\right) \\ $$$${plus}\:{that}\:{we}\:{have}\:{p}^{''} \left({x}\right)={n}\left({n}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{{n}−\mathrm{2}} \:\Rightarrow{p}^{''} \left({p}\mathrm{0}\right)={n}\left({n}−\mathrm{1}\right) \\ $$$${p}^{''} \left(\mathrm{0}\right)\neq\mathrm{0}\:\Rightarrow\mathrm{0}\:{is}\:{root}\:{of}\:{p}\left({x}\right)\:{ar}\:{ordre}\:\mathrm{2}. \\ $$$$ \\ $$
Commented by abdo imad last updated on 09/Mar/18
your answer is also correct sir rachid...
$${your}\:{answer}\:{is}\:{also}\:{correct}\:{sir}\:{rachid}… \\ $$

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