WAŻNE!!!
1. Wykaż że ciąg (tg30, tg45, tg60) jest geometryczny
2. Ciąg geometryczny an określony jest wzorem an=. Oblicz iloraz tego ciągu
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z.1
a1 = tg 30 st = 1/ p(3)
a2 = tg 45 st = 1
a3 = tg 60 st = p(3)
Mamy
a2 : a1 = 1 / ( 1/p(3)) = p(3)
oraz
a3 : a2 = p(3) / 1 = p(3)
zatem q = p(3)
Ciag jest geometryczny.
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p(3) - pierwiastek kwadratowy z 3
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z.2
a1 = [ p(3)]^1 + [ p(3)]^3 = p(3) + 3 p(3) = 4 p(3)
a2 = [ p(3)]^2 + [ p(3)]^4 = 3 + 9 = 12
a3 = [ p(3)]^3 + [ p(3)]^5 = 3 p(3) + 9 p(3) = 12 p(3)
mamy
a2 : a1 = 12 / 4 p(3) = 3/ p(3) = p(3)
oraz
a3 : a2 = 12 p(3) / 12 = p(3)
zatem
q = p(3)
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