المتتاليات العددية
Author:
khaled
Last Updated:
10 yıl önce
License:
Creative Commons CC BY 4.0
Abstract:
تمارين حول المتتاليات العددية
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\documentclass[a4paper,12pt]{article}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=1.7cm,headsep=5mm]{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{enumitem}
\usepackage{fancyhdr}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{siunitx}
\usepackage{fourier} % math font
\usepackage[normalem]{ulem}
\usepackage{esvect}
\usepackage{polyglossia}
% desactiver message d'erreur bidi package -------------------
\makeatletter
\AtBeginDocument{\bidi@isloaded[]{arabxetex}}
\makeatother
% languages & fonts===========================================
\setdefaultlanguage[calendar=gregorian,numerals=maghrib]{arabic}
\setotherlanguage{english}
\newfontfamily\arabicfont[Script=Arabic,Scale=1.1]{Amiri}
\newfontfamily\arabicfontsf[Script=Arabic,Scale=1.2]{Dimnah}
%=============================================================
\usepackage[novoc]{arabxetex}
% Pour permettre à polyglosssia et arabxetex de coexister:
\let\aemph\veryundefinedcommand
\pagestyle{fancy}
\renewcommand{\headrulewidth}{.7pt}
\renewcommand{\footrulewidth}{.7pt}
\rhead{شعبة العلوم التجريبية}
\chead{}
\lhead{ }
\cfoot{\thepage/3}
\renewcommand{\labelenumi}{\textLR{\arabic{enumi}-}}
\renewcommand{\labelenumii}{\alph{enumii})}
\renewcommand\uline{\bgroup\markoverwith%
{%
\rule[-1ex]{2pt}{0.7pt}%
%
}%
\ULon}
\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}