a)
Vi kjenner a 1 = − 3 a_{1}=-3 a 1 = − 3 a n = 69 a_{n}=69 a n = 69 n n n
a n = a 1 + ( n − 1 ) ⋅ d 69 = − 3 + ( n − 1 ) ⋅ 3 69 3 = − 3 + ( n − 1 ) ⋅ 3 3 23 = − 1 + ( n − 1 ) 23 + 1 + 1 = n n = 25 \begin{aligned}
a_{n}&=a_{1}+(n-1)\cdot d \\
69&=-3 + (n-1)\cdot 3 \\
\frac{69}{3} &=\frac{-\cancel{ 3 }+(n-1)\cdot \cancel{ 3 }}{\cancel{ 3 }} \\
23&=-1+(n-1) \\
23 + 1+1&=n \\
n&=25
\end{aligned} a n 69 3 69 23 23 + 1 + 1 n = a 1 + ( n − 1 ) ⋅ d = − 3 + ( n − 1 ) ⋅ 3 = 3 − 3 + ( n − 1 ) ⋅ 3 = − 1 + ( n − 1 ) = n = 25 Summen av den aritmetisk rekka er dermed
s n = a 1 + a n 2 ⋅ n = − 3 + 69 2 ⋅ 25 = 66 2 ⋅ 25 = 33 ⋅ 25 = 825 ‾ ‾ s_{n}=\frac{a_{1}+a_{n}}{2}\cdot n =\frac{-3+69}{2}\cdot 25=\frac{66}{2}\cdot 25=33 \cdot 25 = \underline{\underline{825}} s n = 2 a 1 + a n ⋅ n = 2 − 3 + 69 ⋅ 25 = 2 66 ⋅ 25 = 33 ⋅ 25 = 825
b)
Konvergensområdet er de verdiene av x x x − 1 < k ( x ) < 1 -1< k(x)<1 − 1 < k ( x ) < 1 k ( x ) = 1 2 − x k(x)=\frac{1}{2}-x k ( x ) = 2 1 − x
− 1 < k ( x ) < 1 − 1 < 1 2 − x < 1 1 > − 1 2 + x > − 1 1 + 1 2 > − 1 2 + x + 1 2 > − 1 + 1 2 3 2 > x > − 1 2 \begin{aligned}
-1&<k(x)<1 \\
-1&< \frac{1}{2} -x < 1 \\
1 &> -\frac{1}{2} + x > -1 \\
1 + \frac{1}{2} &> -\cancel{ \frac{1}{2} } + x + \cancel{ \frac{1}{2} } > -1 + \frac{1}{2}\\
\frac{3}{2} &> x > -\frac{1}{2}
\end{aligned} − 1 − 1 1 1 + 2 1 2 3 < k ( x ) < 1 < 2 1 − x < 1 > − 2 1 + x > − 1 > − 2 1 + x + 2 1 > − 1 + 2 1 > x > − 2 1 Konvergensområdet for rekka er x ∈ ⟨ − 1 2 , 3 2 ⟩ ‾ ‾ \underline{\underline{x \in \left\langle-\frac{1}{2}, \frac{3}{2} \right\rangle}} x ∈ ⟨ − 2 1 , 2 3 ⟩
c)
Ballen vil bevege seg på følgende måte:
2 2 2 2 ⋅ 0,75 = 1,5 2 \cdot 0{,}75=1{,}5 2 ⋅ 0 , 75 = 1 , 5 2 ⋅ 0,75 = 1,5 2\cdot 0{,}75=1{,}5 2 ⋅ 0 , 75 = 1 , 5 1,5 ⋅ 0,75 = 1,125 1{,}5 \cdot 0{,}75 =1{,}125 1 , 5 ⋅ 0 , 75 = 1 , 125 1,5 ⋅ 0,75 = 1,125 1{,}5 \cdot 0{,}75 =1{,}125 1 , 5 ⋅ 0 , 75 = 1 , 125 Og så videre …
Ballens totale distanse kan altså modelleres ved hjelp av to geometriske rekker, a a a b b b k = 0,75 k=0{,}75 k = 0 , 75 a 1 = 2 a_{1}=2 a 1 = 2 b 1 = 1,5 b_{1}=1{,}5 b 1 = 1 , 5
s a = a 1 1 − k = 2 1 − 3 4 = 2 1 4 = 2 ⋅ 4 1 4 ⋅ 4 = 8 1 = 8 s b = b 1 1 − k = 1,5 1 − 3 4 = 1,5 1 4 = 1,5 ⋅ 4 1 4 ⋅ 4 = 6 1 = 6 \begin{aligned}
s_{a}&=\frac{a_{1}}{1-k}=\frac{2}{1-\frac{3}{4}}=\frac{2}{\frac{1}{4}}=\frac{2 \cdot 4}{\frac{1}{4}\cdot 4}=\frac{8}{1}=8 \\
s_{b}&=\frac{b_{1}}{1-k}=\frac{1{,}5}{1-\frac{3}{4}}=\frac{1{,}5}{\frac{1}{4}}=\frac{1{,}5 \cdot 4}{\frac{1}{4}\cdot 4}=\frac{6}{1}=6
\end{aligned} s a s b = 1 − k a 1 = 1 − 4 3 2 = 4 1 2 = 4 1 ⋅ 4 2 ⋅ 4 = 1 8 = 8 = 1 − k b 1 = 1 − 4 3 1 , 5 = 4 1 1 , 5 = 4 1 ⋅ 4 1 , 5 ⋅ 4 = 1 6 = 6 Ballen vil totalt bevege seg 14 meter.