Induksjon
Lagt inn: 21/05-2017 13:59
vi setter
[tex]S_n=\frac{2}{1*3}+\frac{2}{2*4}+\frac{2}{3*5}+...+\frac{2}{n(n+2)}, n\geq 1[/tex]
vis ved induksjon at
[tex]S_n=\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}[/tex]
Trinn 1: test for n=1
VS : [tex]\frac{2}{1*3}=\frac{2}{3}[/tex] HS: [tex]\frac{3}{2}-\frac{1}{1+1}-\frac{1}{1+2}=\frac{2}{3}[/tex]
Trinn 2: antar at det stemmer for n=t
[tex]S_t=\frac{2}{1*3}+\frac{2}{2*4}+...+\frac{2}{t(t+2)}=\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}[/tex]
Skal vise at det stemmer for n=t+1
Skal altså vise at :
[tex]S_{t+1}=\frac{2}{1*3}+...+\frac{2}{t(t+2)}+\frac{2}{(t+1)((t+1)+2)}=\frac{3}{2}-\frac{1}{((t+1)+1)}-\frac{1}{(t+1)+2}=\frac{(3t+8)(t+1)}{2(t+2)(t+3)}[/tex]
Bevis:
[tex]S_{t+1}=S_{t}+a_{t+1}=\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}+\frac{2}{(t+1)((t+1)+2)}=\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}+\frac{2}{(t+1)(t+3)}[/tex]
[tex]\left (\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}+\frac{2}{(t+1)(t+3)} \right )*FN[/tex]
[tex]\frac{3(t+1)(t+2)(t+3)}{FN}-\frac{1*2(t+2)(t+3)}{FN}-\frac{1*2(t+1)(t+3)}{FN}+\frac{2*2(t+2)}{FN}=\frac{(t+1)^2(3t+8)}{FN}[/tex]
[tex]FN=2(t+1)(t+2)(t+3)[/tex]
[tex]S_{t+1}=\frac{(t+1)^2(3t+8)}{2(t+1)(t+2)(t+3)}=\frac{(t+1)(3t+8)}{2(t+2)(t+3)}[/tex]
er dette riktig?
[tex]S_n=\frac{2}{1*3}+\frac{2}{2*4}+\frac{2}{3*5}+...+\frac{2}{n(n+2)}, n\geq 1[/tex]
vis ved induksjon at
[tex]S_n=\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}[/tex]
Trinn 1: test for n=1
VS : [tex]\frac{2}{1*3}=\frac{2}{3}[/tex] HS: [tex]\frac{3}{2}-\frac{1}{1+1}-\frac{1}{1+2}=\frac{2}{3}[/tex]
Trinn 2: antar at det stemmer for n=t
[tex]S_t=\frac{2}{1*3}+\frac{2}{2*4}+...+\frac{2}{t(t+2)}=\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}[/tex]
Skal vise at det stemmer for n=t+1
Skal altså vise at :
[tex]S_{t+1}=\frac{2}{1*3}+...+\frac{2}{t(t+2)}+\frac{2}{(t+1)((t+1)+2)}=\frac{3}{2}-\frac{1}{((t+1)+1)}-\frac{1}{(t+1)+2}=\frac{(3t+8)(t+1)}{2(t+2)(t+3)}[/tex]
Bevis:
[tex]S_{t+1}=S_{t}+a_{t+1}=\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}+\frac{2}{(t+1)((t+1)+2)}=\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}+\frac{2}{(t+1)(t+3)}[/tex]
[tex]\left (\frac{3}{2}-\frac{1}{t+1}-\frac{1}{t+2}+\frac{2}{(t+1)(t+3)} \right )*FN[/tex]
[tex]\frac{3(t+1)(t+2)(t+3)}{FN}-\frac{1*2(t+2)(t+3)}{FN}-\frac{1*2(t+1)(t+3)}{FN}+\frac{2*2(t+2)}{FN}=\frac{(t+1)^2(3t+8)}{FN}[/tex]
[tex]FN=2(t+1)(t+2)(t+3)[/tex]
[tex]S_{t+1}=\frac{(t+1)^2(3t+8)}{2(t+1)(t+2)(t+3)}=\frac{(t+1)(3t+8)}{2(t+2)(t+3)}[/tex]
er dette riktig?