Let {a
n} be a sequence of numbers that satisfy a
0 = 3, and (3 − a
n+1)(6 + a
n) = 18. Find
n 1
∑ ---
i=0 ai
Start with the equation (3-a(n+1)) * (6+a(n)) = 18
18 - 6*a(n+1) + 3*a(n) - a(n+1)*a(n) = 18
6*a(n+1) - 3*a(n) + a(n+1)*a(n) = 0
6/a(n) - 3/a(n+1) + 1 = 0
1/a(n+1) = 2/a(n) + 1/3
This last equation is a simple linear recurrence for 1/a(n), and we know 1/a(0) = 1/3. For clarity let b(n) = 1/a(n). Then b(n+1) = 2*b(n) + 1/3 and the goal is the sum of b(0) through b(n).
A little exploration: evaluate the terms of b(n):
b(0) = 1/3
b(1) = 2*(1/3) + 1/3 = 3/3
b(2) = 2*(3/3) + 1/3 = 7/3
b(3) = 2*(7/3) + 1/3 = 15/3
At this point the formula for b(n) is apparent: b(n) = (2^(n+1) - 1)/3. Can be proven more formally with a inductive proof.
Then the sum of the first n+1 terms, starting at n=0 can be written as (2^1 - 1)/3 + (2^2 - 1)/3 + ... + (2^(n+1) - 1)/3
= (2^1 + 2^2 + ... + 2^(n+1))/3 - (1+1+..+1)/3
= (2^(n+2) - 2)/3 - (n+1)/3
= (2^(n+2) - n - 3)/3
No computers needed!
Solution
