Odgovor:
a)# x = 2 #
b) glej spodaj
Pojasnilo:
a) Ker so prvi trije izrazi #sqrt x-1 #, 1 in #sqrt x + 1 #srednji izraz, 1, mora biti geometrična sredina drugih dveh. Zato
# 1 ^ 2 = (sqrt x-1) (sqrt x +1) pomeni #
# 1 = x-1 pomeni x = 2 #
b)
Takrat je skupno razmerje #sqrt 2 + 1 #in prvi mandat je #sqrt 2-1 #.
Tako je peti mandat
# (sqrt 2-1) krat (sqrt 2 + 1) ^ 4 = (sqrt 2 + 1) ^ 3 #
#qquad = (sqrt 2) ^ 3 + 3 (sqrt2) ^ 2 + 3 (sqrt2) + 1 #
# qquad = 2sqrt2 + 6 + 3sqrt2 + 1 #
#qquad = 7 + 5sqrt2 #
Odgovor:
Glej spodaj.
Pojasnilo:
Glede na to, # rarrsqrtx-1,1, sqrtx + 1 # so v # GP #.
Torej, #rarr (sqrtx-1) / 1 = 1 / (sqrtx + 1) #
#rarr (sqrtx-1) ^ 2 = 1 #
#rarr (sqrtx) ^ 2-1 ^ 2 = 1 #
# rarrx = 2 #
Prvi mandat # (a) = sqrtx-1 = sqrt2-1 #
Drugi mandat # (b) = 1 #
Skupno razmerje # (r) = b / a = 1 / (sqrt2-1) = sqrt2 + 1 #
The # n ^ (th) # izraz geometrijskega zaporedja # (t_n) = a * r ^ (n-1) #
Torej, # t_5 = (sqrt2-1) * (sqrt2 + 1) ^ (5-1) #
# = (sqrt2-1) (sqrt2 + 1) (sqrt2 + 1) ^ 3 #
# = (sqrt2) ^ 2-1 ^ 2 (sqrt2) ^ 3 + 3 * (sqrt2 ^ 2) * 1 + 3 * sqrt2 * 1 ^ 2 + 1 ^ 3 #
# = (2-1) (2sqrt2 + 6 + 3sqrt2 + 1) = 7 + 5sqrt2 #
Odgovor:
# x = 2 in 5 ^ (th) "izraz" = 7 + 5sqrt2 #.
Pojasnilo:
Za kaj #3# zaporednih izrazov # a, b, c # a GP, imamo, # b ^ 2 = ac #.
Zato v našem primeru # 1 ^ 2 = (sqrtx-1) (sqrtx + 1) = (sqrtx) ^ 2-1 ^ 2, #
# t.j. 1 = x-1 ali x = 2 #.
S # x = 2 #, # 1 ^ (st) in 2 ^ (nd) # pogoji GP Spodaj
sklic, # sqrtx-1 = sqrt2-1 in 1 #, resp.
Torej skupno razmerje # r = (2 ^ (nd) "izraz) -:(1 ^ (st)" izraz) "#, # = 1 / (sqrt2-1) = sqrt2 + 1 #.
#:. 4 ^ (th) "izraz = r (" 3 ^ (rd) "izraz) = (sqrt2 + 1) (sqrtx + 1) #, # = (sqrt2 + 1) (sqrt2 + 1) #, # = 2 + 2sqrt2 + 1 #, # = 3 + 2sqrt2 #.
Nadalje, # (5 ^ (th) "izraz) = r (" 4 ^ (th) izraz) #, # = (sqrt2 + 1) (3 + 2sqrt2) #,
# = 3sqrt2 + 3 + 2sqrt2 * sqrt2 + 2sqrt2 #.
# rArr 5 ^ (th) "izraz" = 7 + 5sqrt2 #.
