Expert: Sherman D. - 9/3/2007

Question
Give the first six terms of the sequence defined by:

a1=1   a(k+1)=1+[1/a(k)] for k>=1

Answer
a(k + 1) = 1 + (1/(a(k)))

a(k) = 1 + d(k - 1)
a(k) = 1 + dk - d
dk = a(k) - 1 + d

a(k + 1) = 1 + d((k + 1) - 1)
a(k + 1) = 1 + dk
a(k + 1) = 1 + a(k) - 1 + d
a(k + 1) = a(k) + d

1 + (1/(a(k))) = a(k) + d
(a(k) + 1)/(a(k)) = a(k) + d
a(k) + 1 = (a(k))^2 + d(a(k))
(a(k))^2 + (d - 1)a(k) - 1 = 0

so for this to work (d - 1) = 0, so d = 1

-(1/(a(k)) = 1 - a(k + 1)
(1/(a(k))) = a(k + 1) - 1
a(k) = 1/(a(k + 1) - 1)

a(2) = 1 + (2 - 1) = 1 + 1 = 2
a(3) = 1 + (3 - 1) = 1 + 2 = 3
a(4) = 4
a(5) = 5
a(6) = 6

a(k) = 1/(a(k + 1) - 1)

a(2) = 1/(2 - 1) = 1/1
a(3) = 1/2
a(4) = 1/3
a(5) = 1/4
a(6) = 1/5

ANS : 1, 1, (1/2), (1/3), (1/4), (1/5)

not completely certain. Check with answers.yahoo.com to see if that is what they get, or if they can show you an easier way.