Solve the recurrence an 2 a2n a3n 1 with a1 1 and a2 2S
Solve the recurrence an + 2 a2n = a3n + 1 with a1 = 1 and a2 = 2.
Solution
Given that an+2an2 = a3n+1
Take log on both sides
ln an+2 + 2ln an = 3 ln an+1
Or ln an+2 = -2ln an + 3ln an+2
Let n =1
ln a3 = -2 ln 1+3ln 2 = 3ln 2
a3 =8
lna4 = -2ln2+3ln 8
a4 = 128
ln a5 = -2ln8+3ln 128
a5 = 128^3/64 = 32768 and so on.
In general
ln an+2 = 3 ln an+1-2ln an
= 3(3ln an-2ln an-1) -2ln an
and so on.
