Question

# 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

Solution:

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 $47\sum_{n=1}^{\infty}\frac{a_n}{8^n} = 7$