Question

Let
\(\{a_n\}_{n=1}^{\infty}\)
be a sequence such that \(a_1 = 1\), \(a_2 = 1\) and
\(a_{n+2} = 2a_{n+1} + a_n\)
for all \(n \ge 1\). Then the value of
\(47\sum\limits_{n=1}^{\infty} \frac{a_n}{2^{3n}}\)
is equal to ______________.

Hint:

Let \(S = \sum a_n/8^n\) and use the recurrence to get a linear equation in S.

Answer 1

Correct Answer: 7

Let \(S = \sum_{n \ge 1} \frac{a_n}{8^n}\). Split off the first two terms and apply the recurrence:
\(S = \frac{a_1}{8} + \frac{a_2}{64} + \sum_{n \ge 1}\frac{2a_{n+1} + a_n}{8^{n+2}}\)
\(= \frac{1}{8} + \frac{1}{64} + \frac{2}{8}\left(S - \frac{a_1}{8}\right) + \frac{S}{64}\) (using \(\sum_{n\ge1} \frac{a_{n+1}}{8^{n+1}} = S - \frac{a_1}{8}\) and \(\sum_{n \ge 1} \frac{a_n}{8^{n}} = S\)).
\(S = \frac{9}{64} + \frac{S}{4} - \frac{1}{32} + \frac{S}{64} \Rightarrow S\left(1 - \frac{1}{4} - \frac{1}{64}\right) = \frac{7}{64} \Rightarrow \frac{47}{64}S = \frac{7}{64}\).
So \(S = \frac{7}{47}\) and \(47S = 7\).
Answer: 7