اكتب كلاً من المتتابعات الآتية مكتفياً بذكر الحدود السنة الأولى: ${\mathrm{u}}_{\mathrm{n}}=(-1{)}^2$

$$\begin{gathered} \begin{array}{rl}& {\mathrm{u}}_1=(-1{)}^1=-1 , {\mathrm{u}}_2=(-1{)}^2=1\\ & {\mathrm{u}}_3=(-1{)}^3=-1\\ & {\mathrm{u}}_4=(-1{)}^4=1 , {\mathrm{u}}_5=(-1{)}^5=-1 , {\mathrm{u}}_6=(-1{)}^6=1\end{array} \\ ⟨{\mathrm{u}}_{\mathrm{n}}⟩=⟨-1,1,-1,1,-1,1,\dots \dots ⟩ \end{gathered}$$

تمارين (1-2)

(1)- أي العبارات التالية صحيحة وأيها خاطئة:

1. كل دالة مجالها ⁺Z هي متتابعة. (صح)
2. كل دالة مداها ⁺Z هي متتابعة. (خطأ)
3. كل دالة مجالها {3,4,5,6,7,8} هي متتابعة. (خطأ)
4. كل دالة مجالها Z هي متتابعة. (خطأ)
5. كل دالة مجالها {1,2,3,4,5,6,7} متتابعة منتهية. (صح)
6. كل دالة مجالها {1,2,3,4,5,6,7,8,9} هي متتابعة. (خطأ)
7. الحد الرابع في المتتابعة $⟨\sqrt{\mathrm{n}}/\mathrm{n}+1⟩$ يساوي 2/5. (صح)
8. مجال المتتابعة <96 , ... , 2,4,6> هو ⁺Z. (خطأ)
9. في المتتابعة <U_n> حيث U_n-1=n U_n
10. فإن الحدان الأول والثاني مختلفان عندما 1=n. (خطأ)
11. في المتتابعة <n²> يكون Un+1 < U_n. (خطأ)

(2)- اكتب كلاً من المتتابعات الآتية مكتفياً بذكر الحدود السنة الأولى:

أ- ${\mathrm{u}}_{\mathrm{n}}={\mathrm{n}}^2-2\mathrm{n}$

$$\begin{gathered} {\mathrm{u}}_1=1^2-2\times 1=1-2=-1 , {\mathrm{u}}_2=2^2-2\times 2=4-4=\underset{}{0} \\ {\mathrm{u}}_1=3^2-2\times 3=\underset{}{3 }, {\mathrm{u}}_4=4^2-2\times 4=8 \\ {\mathrm{u}}_5=5^2-2\times 5=15 ,\hspace{0.17em}{\mathrm{u}}_6=6^2-2\times 6=36-12=24 \\ <{\mathrm{u}}_{\mathrm{n}}>=<-1,0,3,8,15,24\text{,...}> \end{gathered}$$

ب- ${\mathrm{u}}_{\mathrm{n}}=2$

$<{\mathrm{u}}_{\mathrm{n}}>=<2,2,2,2,2,2,...>$

ج- ${\mathrm{u}}_{\mathrm{n}}=\frac{6}{\mathrm{n}}$

$\begin{array}{rl}{\mathrm{u}}_1& =\frac{6}{1}=\underset{}{6 }, {\mathrm{u}}_2=\frac{6}{2}=3 , {\mathrm{u}}_3=\frac{6}{3}=2 , {\mathrm{u}}_4=\frac{6}{4}=\frac{3}{2} , {\mathrm{u}}_5=\frac{6}{5} , {\mathrm{u}}_6=\frac{6}{6}=1\\ & <{\mathrm{u}}_{\mathrm{n}}>=<6,3,2,\frac{3}{2},\frac{6}{5},1,\dots .>\end{array}$

د- ${\mathrm{u}}_{\mathrm{n}+1}=\frac{4}{1+{\mathrm{u}}_{\mathrm{n}}} , {\mathrm{u}}_1=1$

$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{n}+1}=\frac{4}{1+{\mathrm{u}}_{\mathrm{n}}}\to \mathrm{n}+1-1=\mathrm{n}\\ & {\mathrm{u}}_2=\frac{4}{1+{\mathrm{u}}_1}=\frac{4}{1+1}=\frac{4}{2}=2 , {\mathrm{u}}_3=\frac{4}{1+{\mathrm{u}}_2}=\frac{4}{1+2}=\frac{4}{\underset{}{3}}\\ & {\mathrm{u}}_4=\frac{4}{1+{\mathrm{u}}_3}=\frac{4}{1+\frac{4}{3}}=\frac{4}{7}=4\cdot \frac{3}{7}=\frac{12}{7}\\ & {\mathrm{u}}_5=\frac{4}{1+{\mathrm{u}}_4}=\frac{4}{1+\frac{12}{7}}=\frac{4}{19}=\frac{4\cdot 7}{19}=\frac{28}{19}\\ & {\mathrm{u}}_6=\frac{4}{47}=\frac{4\cdot 19}{47}=\frac{76}{47}\\ & <{\mathrm{u}}_{\mathrm{n}}>=<1,2,\frac{4}{3},\frac{12}{7},\frac{28}{19},\frac{76}{47},\dots \dots >\end{array}$

هـ- ${\mathrm{u}}_{\mathrm{n}}=1-\frac{2}{\mathrm{n}}$

$$\begin{gathered} {\mathrm{u}}_1=1-\frac{2}{1}=-1 , {\mathrm{u}}_2=1-\frac{2}{2}=0,{\mathrm{u}}_3=1-\frac{2}{3}=\frac{1}{3} \\ \begin{array}{rl}& {\mathrm{u}}_4=1-\frac{2}{4}=1-\frac{1}{2}=\frac{1}{2}\\ & {\mathrm{u}}_5=1-\frac{2}{5}=\frac{3}{5}-\frac{{\mathrm{u}}_2}{2}\\ & {\mathrm{u}}_6=1-\frac{2}{6}=1-\frac{1}{3}=\frac{2}{3}\\ & <{\mathrm{u}}_{\mathrm{n}}>=<-1,0,\frac{1}{3},\frac{1}{2},\frac{3}{5},\frac{2}{3},\dots \dots .>\end{array} \end{gathered}$$

و- ${\mathrm{u}}_{\mathrm{n}}=(-1{)}^2$

$$\begin{gathered} \begin{array}{rl}& {\mathrm{u}}_1=(-1{)}^1=-1 , {\mathrm{u}}_2=(-1{)}^2=1\\ & {\mathrm{u}}_3=(-1{)}^3=-1\\ & {\mathrm{u}}_4=(-1{)}^4=1 , {\mathrm{u}}_5=(-1{)}^5=-1 , {\mathrm{u}}_6=(-1{)}^6=1\end{array} \\ ⟨{\mathrm{u}}_{\mathrm{n}}⟩=⟨-1,1,-1,1,-1,1,\dots \dots ⟩ \end{gathered}$$

ز- ${\mathrm{u}}_{\mathrm{n}}=2^{\mathrm{n}-1}$

$\begin{array}{rl}{\mathrm{u}}_1& =2^{1-1}=2^0=1 , {\mathrm{u}}_2=2^{2-1}=2^1=2 , {\mathrm{u}}_3=2^{3-1}=2^2=4\\ {\mathrm{u}}_4& =2^{4-1}=2^3=8 , {\mathrm{u}}_5=2^{5-1}=2^4=\underset{}{16 }, {\mathrm{u}}_6=2^{6-1}=2^5=\underset{}{32}\\ & <{\mathrm{u}}_{\mathrm{n}}>=⟨1,2,4,8,16,32,\dots \dots >\end{array}$

ح- ${\mathrm{U}}_{\mathrm{n}}=\{\begin{array}{c}1\dots .\mathrm{odd} \mathrm{n}\\ 2\dots \mathrm{even} \mathrm{n}\end{array}$

$\text{Domain}=\{1,2,\underset{\_}{3},4,\underset{\_}{5},\underset{\_}{6},\underset{\_}{7},\underset{\_}{8},\dots \dots \dots \}$

• الفردية: ${\mathrm{u}}_1=1 , {\mathrm{u}}_3=1 , {\mathrm{u}}_5=1 , {\mathrm{u}}_7=1$
• الزوجية: ${\mathrm{u}}_2=2 , {\mathrm{u}}_4=2 , {\mathrm{u}}_6=2 , {\mathrm{u}}_8=2$

$⟨{\mathrm{u}}_{\mathrm{n}}⟩=⟨1,2,1,2,1,2,\dots \dots ⟩$

(3)- في المتتابعة <Un> حيث Un= n²+2n أثبت أن Un+1 > Un

في المعادلة U_n= n²+2 مرة نعوض عن n ب n ومرة أخرى عن n ب n+1

$\begin{array}{rl}& \mathrm{n}=\mathrm{n},{\mathrm{u}}_{\mathrm{n}}={\mathrm{n}}^2+2\mathrm{n}\\ & \mathrm{n}=\mathrm{n}+1,{\mathrm{u}}_{\mathrm{n}+1}=(\mathrm{n}+1{)}^2+2(\mathrm{n}+1)\end{array}$

البرهان:

$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{n}+1}>{\mathrm{u}}_{\mathrm{n}} \text{mean }{\mathrm{u}}_{\mathrm{n}+1}-{\mathrm{u}}_{\mathrm{n}}>0\\ & (\mathrm{n}+1{)}^2+2(\mathrm{n}+1)-({\mathrm{n}}^2+2\mathrm{n})>0\\ & (\mathrm{n}+1{)}^2+2\mathrm{n}+2-{\mathrm{n}}^2-2\mathrm{n}>0\\ & {\mathrm{n}}^2+2\mathrm{n}+1+2\mathrm{n}+2-{\mathrm{n}}^2-2\mathrm{n}>0\\ & 2\mathrm{n}+3>0\end{array}$

(4)- اكتب ثمانية حدود من المتتابعة بفرض:

$\Rightarrow \mathrm{U}:{\mathrm{Z}}^{+}\to \mathrm{R},{\mathrm{U}}_{\mathrm{n}}=\left\{\begin{array}{l}\mathrm{n}+2\dots \mathrm{odd} \mathrm{n}\\ \frac{4}{\mathrm{n}}\cdots \mathrm{even} \mathrm{n}\end{array}\right.$

• الفردية: ${\mathrm{u}}_{\mathrm{n}}=\mathrm{n}+2 , {\mathrm{u}}_1=1+2=\underset{}{3} , {\mathrm{u}}_3=3+2=\underset{}{5 }, {\mathrm{u}}_5=5+2=\underset{}{7 }, {\mathrm{u}}_7=7+2=\underset{}{9}$
• الزوجية: ${\mathrm{u}}_{\mathrm{n}}=\frac{4}{\mathrm{n}} ,\hspace{0.17em}{\mathrm{u}}_2=\frac{4}{2}=\underset{}{2 }, {\mathrm{u}}_4=\frac{4}{4}=\underset{}{1 }, {\mathrm{u}}_6=\frac{4}{6}=\frac{2}{\underset{}{3}} , {\mathrm{u}}_8=\frac{4}{8}=\frac{1}{\underset{}{2}}$

$<{\mathrm{u}}_{\mathrm{n}}>=\{<3,2,5,1,7,\frac{2}{3},9,\frac{1}{2},\dots >\}$