المتتاليات العددية
Author:
khaled
Last Updated:
10 лет назад
License:
Creative Commons CC BY 4.0
Аннотация:
تمارين حول المتتاليات العددية
\begin
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\rhead{شعبة العلوم التجريبية}
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\begin{document}
\centerline{\Large\bf سلسلة تمارين في المتتاليات العددية }\vspace{0mm}
\rule{\textwidth}{1pt}\\
اعداد: حريز خالد\hfill \\
\rule{\textwidth}{1pt}\\
\subsubsection*{\uline{\sffamily{\large التمرين الأول:}}}
\begin{arab}
almtataalyT al`dadyT $ (u\sb{n}) $ m`rfT kmaa yly:
$ \left\lbrace\begin{array}{ll}
u\sb{0}&=1\\
u\sb{n+1}&=\dfrac{1}{2}u\sb{n}-1
\end{array} \right. \quad \forall n\in\mathbb{N}$
\begin{enumerate}
\item 'a.hsb $ u\sb{3},u\sb{2},u\sb{1} $.
\begin{enumerate}
\item 'a_tbt baaltraaj` 'anh mn 'ajl kl `dad.tby`y $n$: $u\sb{n}\geq -2$.
\item jd 'aitjaah t.gyr almtataalyT $ (u\sb{n}) $. maa_dA tstntj ?
\end{enumerate}
\item $ (v\sb{n} )$almtatAlyT al`dadyT alm`rfT mn 'ajl kl `dad.tby`y $n$
kmA yly: $v\sb{n}=u\sb{n}+2$.
\begin{enumerate}
\item byn 'an almtatAlyT $(v\sb{n})$ hndsyT.
\item `br bdlaalT $n$ `n al.hd al`Am $v\sb{n}$ _tm $u\sb{n}$.
\item 'a.hsb $\lim\limits\sb{n\rightarrow +\infty}u\sb{n}$.
\item 'a.hsb bdlaalT $n$ almjmw` $S\sb{n}$.hy_t:
$S\sb{n}=u\sb{0}+u\sb{1}+\ldots +u\sb{n}$
\end{enumerate}
\end{enumerate}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين الثاني:}}}
\begin{arab}
\begin{enumerate}
\item n`tbr almtatAlyT al`dadyT $(u\sb{n})$ alm`rfT b--:
$ \left\lbrace\begin{array}{lcl}
u\sb{0}&=&-1\\
3u\sb{n+1}&=&u\sb{n}+4 \quad \forall n\in\mathbb{N}
\end{array} \right. $
\begin{enumerate}
\item brhn baaltrAj` 'anh mn 'ajl kl `dad.tby`y $n$ ykwn: $u\sb{n}\leq 2$.
\item byn 'an almtataalyT $(u\sb{n})$ mtzAydT.
\item 'istantj m` altbryr 'an almtatAlT $(u\sb{n})$ mtqArbT.
\end{enumerate}
\item n.d` mn 'ajl kl `dad.tby`y $n$: $v\sb{n}=u\sb{n}-2$.
\begin{enumerate}
\item byn 'an $(v\sb{n})$ mtataalyT hndsyT y.tlb t`yeyn 'saashaa w
.hdhaa al'awl$v\sb{0}$.
\item 'aktb al.hd al`Am $v\sb{n}$ bdlAlT $n$ _tm 'istantj al.hd al`Am
$u\sb{n}$ bdlaalT $n$.
\item 'a.hsb $\lim\limits\sb{n\rightarrow +\infty}u\sb{n}$.
\item 'a.hsb bdlaalT $n$ almjmw` $S\sb{n}$.hy_t:
$S\sb{n}=u\sb{0}+u\sb{1}+\ldots +u\sb{n}$.\\
- `yn al`dad al.tby`y $n$ b.hy_t ykwn:$S\sb{n}=28-3n$.
\end{enumerate}
\end{enumerate}
\newpage
\rhead{}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين الثالث:}}}
\begin{arab}
$ (u\sb{n}) $ mtatAlyT `dadyT m`rfT kmA yly:
$ \left\lbrace\begin{array}{lcl}
u\sb{0}&=&\alpha ,\qquad (\alpha\in\mathbb{R})\\
u\sb{n+1}&=&\dfrac{2}{3}u\sb{n}-\dfrac{8}{9} ,\qquad (\forall n\in\mathbb{N})
\end{array} \right. $
\begin{enumerate}
\item brhn baaltrAj` 'anh fy.hAlT $\alpha =-\dfrac{8}{3}$ tkwn
almtatAlyT $(u\sb{n})$ _tAbtT.
\item fy kl mA yly $\alpha =2$, n`rf almtatAlyT al`dadyT $(v\sb{n})$
kA yly: $v\sb{n}=u\sb{n}+\dfrac{8}{3}$.
\begin{enumerate}
\item 'a.hsb $u\sb{2} , u\sb{1}$.
\item 'a_tbt 'an $(v\sb{n})$ mtatAlyT hndsyT y.tlb t`yeyn 'sAshA $q$ w
.hdhA al'awl $v\sb{0}$.
\item 'aktb `bArT $u\sb{n}$bdlAlT $n$. wA.hsb
$\lim\limits\sb{n\rightarrow +\infty}u\sb{n}$.
\end{enumerate}
\end{enumerate}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين الرابع:}}}
\begin{arab}
ltkn almtatAlyT $ u\sb{n} $alm`rfT kmA yly:
$ \left\lbrace\begin{array}{lcl}
u\sb{1}&=&7\\
u\sb{n+1}&=&\alpha u\sb{n}+5 ,\qquad (\forall n\in\mathbb{N}\sp\ast)
\end{array} \right. $
$\bullet$ n.d` mn 'ajl $v\sb{n}=u\sb{n}-6 ,\quad n\in\mathbb{N}\sp\ast$
\begin{enumerate}
\item 'wjd al`dad al.hqyqy $\alpha$ b.hy_t tkwn $(v\sb{n})$
mtatAlyT hndsyT.\\ w 'a.hsb 'sAshA w.hdhA al'awl fy h_dh al.hAlT.
\item n.d`: $\alpha =\dfrac{1}{6}$.
\begin{enumerate}
\item 'aktb `bArT al.hd al`Am $v\sb{n}$ bdlAlT $n$.
\item 'a.hsb $\lim\limits\sb{n\rightarrow +\infty}v\sb{n}$.
\item 'istntj $\lim\limits\sb{n\rightarrow +\infty}u\sb{n}$.
\end{enumerate}
\item 'a.hsb almjmw` $S\sb{n}=v\sb{0}+v\sb{1}+\ldots +v\sb{n}$ bdlAlT $n$
w 'a.hsb $\lim\limits\sb{n\rightarrow +\infty}S\sb{n}$.
\item 'a.hsb almjmw` $S'\sb{n}=u\sb{0}+u\sb{1}+\ldots +u\sb{n}$ bdlAlT $n$
w 'a.hsb $\lim\limits\sb{n\rightarrow +\infty}S'\sb{n}$.
\end{enumerate}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين الخامس:}}}
\begin{arab}
$ (u\sb{n}) $mtatAlyT m`rfT kmA yly:
$ \left\lbrace\begin{array}{lcl}
u\sb{0}&=&-12\\
u\sb{n+1}&=&\dfrac{3}{4} u\sb{n}-7
\end{array} \right. $
\begin{enumerate}
\item 'a.hsb $u\sb{3} ,u\sb{2},u\sb{1}$.
\item ltkn almtatAlyT $ k\sb{n}=u\sb{n}+\alpha $.
\begin{enumerate}
\item `yn al`dad al.qyqy $\alpha$.htY tkwn $k\sb{n}$
mtatAlyT hndsyT y.tlb t`yeyn 'sAshA w.hdhA al'awl $k\sb{0}$.
\item n.d` $\alpha=28$ 'aktb `bArT $k\sb{n}$ bdlAlT $n$ w 'stntj `bArT
$u\sb{n}$ bdlAlT $n$.
\item brhn `lY.s.hT $u\sb{n}$ bdlAlT $n$ baaltrAj`.
\item 'a.hsb almjmw` $S\sb{n}=u\sb{0}+u\sb{1}+\ldots +u\sb{n}$ bdlAlT $n$
w 'a.hsb $\lim\limits\sb{n\rightarrow +\infty}u\sb{n}$.
\end{enumerate}
\end{enumerate}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين السادس:}}}
\begin{arab}
$ (v\sb{n}) $mtatAlyT m`rfT mn 'ajl kl
`dad.tby`y $n$ kmA yly:
$ \left\lbrace\begin{array}{lcl}
v\sb{0}&=&-\alpha\\
v\sb{n+1}&=& 4v\sb{n}+3
\end{array} \right. $
\begin{enumerate}
\item maa_dA tsmY al`lAqT alm`rfT bhaa almtatAlyT$ (v\sb{n}) $ ?
\item `yn qymT al.hd al'awl $v\sb{0}$ alty tj`l $(v\sb{n})$ _tAbtT.
\item nfr.d fymA yly: 'an $(v\sb{n})$.gyr _tAbtT.hdhA al'awl: $v\sb{0}=5$.
\begin{enumerate}
\item brhn baaltrAj` mn 'ajl kl $n\in\mathbb{N}$ 'an:
jmy`.hdwd $(v\sb{n})$ mwjbT tmAmA. mA_dA tstntj ?
\item 'drs 'itjAh t.gyr $(v\sb{n})$ , hl hy mtqArbT ?
\end{enumerate}
\end{enumerate}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين السابع:}}}
\begin{arab}
ltkn almtatAlyT al`dadyT $(u\sb{n})$
alm`rfT mn 'ajl kl `dad.tby`y $n$:
$ \left\lbrace\begin{array}{lcl}
u\sb{0}&=&1\\
u\sb{n+1}&=&\sqrt{4u\sb{n}}
\end{array} \right. $
\begin{enumerate}
\item brhn baaltraaj` 'anh mn 'ajl kl `dad.tby`y $n$: $u\sb{n}>0$.
\item ltkn $(v\sb{n})$ almtatAlyT alm`rfT kmaa yly: $v\sb{n}=\ln\left(\dfrac{u\sb{n}}{4}\right)$\\
- brhn 'an $(v\sb{n})$ mtatAlyT hndsyT y.tlb t`yiyn 'sAshA w.hdhaa al'awl.\\
- 'a.hsb $v\sb{n}$ bdlaalT $n$ , _tm 'istntj $u\sb{n}$ bdlaalT $n$.\\
- 'a.hsb almjmw` $S\sb{n}=u\sb{0}+u\sb{1}+\ldots +u\sb{n}$ bdlAlT $n$.\\
- 'a.hsb aljdaa' $P\sb{n}=v\sb{0}\times u\sb{1}\times\ldots \times u\sb{n}$ bdlAlT $n$.
\end{enumerate}
\end{arab}
\subsubsection*{\uline{\sffamily{\large التمرين الثامن:}}}
\begin{arab}
n`tbr almtatAlyT $(v\sb{n})$ alm`rfT mn 'ajl kl
$n\in\mathbb{N}$ kmA yly:
$ \left\lbrace\begin{array}{lcl}
v\sb{0}&=&0\\
v\sb{n+1}&=&\dfrac{2}{3}v\sb{n}+\dfrac{1}{3}
\end{array} \right. $
\begin{enumerate}
\item \begin{enumerate}
\item brhn baaltraaj` 'anh mn 'ajl kl `dad.tby`y $n$:
$0\leq v\sb{n}<1$.
\item brhn 'an almtatAlyT $(v\sb{n})$ mtzaaydT tmAmA.
\end{enumerate}
\item ltkn aldAlT $l$alm`rfT `lY$\mathbb{R}$ kmaa yly: $l(x)=\dfrac{2}{3}x+\dfrac{1}{3}$.
\begin{enumerate}
\item `yn al`dad al.hqyqy $l(\alpha)$ b.hy_t ykwn: $l(\alpha)=\alpha$.
\item n.d` mn 'ajl kl `dad.tby`y $n$: $u\sb{n}=v\sb{n}-\alpha$. byn 'an $(u\sb{n})$ mtatAlyT
hndsyT , wA.hsb `bArT $u\sb{n}$ bdlaalT $n$.
\item 'istantj `bArT $v\sb{n}$ bdlaalT$n$ , _tm 'a.hsb
$\lim\limits\sb{n\rightarrow +\infty}v\sb{n}$.
\end{enumerate}
\end{enumerate}
\end{arab}
\lfoot{\uline{وقت ممتع في حل التمارين}}
\end{document}