Let a_n_n=0^ be a sequence such that a_0=a_1=0 and a_n+2 = 2a_n+1-a_n+1 for all n0.

Question

Let $\{a_n\}_{n=0}^{\infty}$ be a sequence such that $a_0=a_1=0$ and $a_{n+2} = 2a_{n+1}-a_n+1$ for all $n\ge0$. Then $\displaystyle\sum_{n=2}^{\infty}\frac{a_n}{7^n}$ is equal to:

Accepted Answer

Computing terms: $a_2=1, a_3=3, a_4=6, a_5=10,\ldots$ (triangular numbers).

Let $s = \sum_{n=2}^\infty \dfrac{a_n}{7^n}$. Applying the successive-subtraction technique twice (multiplying by $\tfrac17$ each time, since $a_n$ has constant second differences):

$$\frac{36s}{49} = \frac{1}{7^2}\cdot\frac{1}{1-\tfrac17} = \frac{7}{49\times6} \implies s = \frac{7}{216}.$$

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