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 n1.

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 $n\ge1$. Then the value of $47\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{2^{3n}}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

Let $x=\sum_{n=1}^{\infty}\dfrac{a_n}{8^n}$. Dividing the recurrence by $8^n$ and summing (a linear recurrence technique), using $a_1=a_2=1$:

$$64x - 9 = 2(8x-1)+x \implies 47x = 7.$$

So $47\displaystyle\sum\frac{a_n}{8^n} = 7$.

Solutions · 1