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La [tex]G_n = 2^{2^n}+1[/tex] for n=0,1,2,...

Vis at

[tex]G_n = 2+\prod_{j=0}^{n-1}G_j [/tex] for [tex]n=1,2,3...[/tex]

[tex]G_{0}=2^{1}+1=3[/tex]

[tex]G_{1}=2^{2^{1}}+1=4+1=5=2+3=2+G_{0}[/tex]

så ok for n=1.

Vi har fylgjande samanheng mellom [tex]G_{n}[/tex] og [tex]G_{n-1}[/tex]:

[tex]G_{n}=(G_{n-1}-1)^2+1[/tex]

Per antakelse veit vi at

[tex]G_{n-1}=2+\displaystyle\prod_{j=0}^{n-2} G_{j}[/tex]

dvs.

[tex]G_{n}= (1+\displaystyle\prod_{j=0}^{n-2} G_{j})^2+1[/tex]

[tex]G_{n}= 2+\displaystyle\prod_{j=0}^{n-2} G_{j}(2+\displaystyle\prod_{j=0}^{n-2} G_{j})=2+(\displaystyle\prod_{j=0}^{n-2} G_{j})\cdot{G_{n-1}}[/tex]

dvs.

[tex]G_{n}=2+\displaystyle\prod_{j=0}^{n-1} G_{j}[/tex]

Ja, riktig!