Let \(\{a_n\}_{n=1}\) be a sequence such that \(a_1 = 1 , a_2 = 1\) and \(a_{n + 2} = 2a_{n + 1} + a_n\) for all \(n \geq 1\). Then the value of \[47 \sum_{n=1}^{\infty} \frac{a_n}{8^n}\] is equal to

Given, \(a_{n+2} = 2a_{n+1} + a_n\)
Dividing both sides by \(8^n\)

Therefore, \(\frac{a_{n+2}}{8^n} = \frac{2a_{n + 1}}{8^n} + \frac{a_n}{8^n}\)
\( \Rightarrow 64 \times \frac{a_{n+2}}{8^{n+2}} = 16 \times \frac{a_{n + 1}}{8^n} + \frac{a_n}{8^n}\)

Applying Summation to both sides

we get, \( 64 \sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = 16 \sum_{n=1}^{\infty}\frac{a_{n + 1}}{8^{n+1}} + \sum_{n=1}^{\infty}\frac{a_n}{8^n}\)

Let \( S = \sum_{n=1}^{\infty}\frac{a_n}{8^n}\)
then \(\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = S – \frac{a_1}{8} – \frac{a_2}{64}\)
\(\sum_{n=1}^{\infty}\frac{a_{n + 1}}{8^{n+1}} = S – \frac{a_1}{8} \)

Therefore from above equation
\(64(S – \frac{a_1}{8} – \frac{a_2}{64}) = 16(S – \frac{a_1}{8} ) + S \)
\( \Rightarrow 64S – 8 – 1 = 17S – 2\)
\( \Rightarrow 47S = 7\)

Therefore \[47\sum_{n=1}^{\infty}\frac{a_n}{8^n} = 7\]