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 $$n \geq 0$$. Then, $$\sum_{n=2}^{\infty} \frac{a_n}{7^n}$$ is equal to

Step 1: Find the pattern of $$a_n$$

Let's compute the first few terms using the initial conditions $$a_0 = 0$$ and $$a_1 = 0$$:

• For $$n = 0$$:$$a_2 = 2a_1 - a_0 + 1 = 2(0) - 0 + 1 = 1$$
• For $$n = 1$$:$$a_3 = 2a_2 - a_1 + 1 = 2(1) - 0 + 1 = 3$$
• For $$n = 2$$:$$a_4 = 2a_3 - a_2 + 1 = 2(3) - 1 + 1 = 6$$
• For $$n = 3$$:$$a_5 = 2a_4 - a_3 + 1 = 2(6) - 3 + 1 = 10$$

The sequence for $$n = 0, 1, 2, 3, 4, 5, \dots$$ is:

$$0, 0, 1, 3, 6, 10, \dots$$

These are the triangular numbers, shifted. Specifically, for $$n \geq 2$$, the general formula is:

$$a_n = \frac{(n-1)n}{2}$$

Step 2: Compute the Infinite Series

We want to evaluate:

$$S = \sum_{n=2}^{\infty} \frac{a_n}{7^n} = \sum_{n=2}^{\infty} \frac{n(n-1)}{2 \cdot 7^n}$$

Let's pull out the constant factor $$\frac{1}{2}$$:

$$S = \frac{1}{2} \sum_{n=2}^{\infty} n(n-1) \left(\frac{1}{7}\right)^n$$

Step 3: Use Derivatives of Geometric Series

Consider the standard infinite geometric series for $$|x| < 1$$:

$$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$

Differentiating with respect to $$x$$ once:

$$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$$

Differentiating a second time:

$$\sum_{n=2}^{\infty} n(n-1) x^{n-2} = \frac{2}{(1-x)^3}$$

Now, multiply both sides by $$x^2$$ to match our series form:

$$\sum_{n=2}^{\infty} n(n-1) x^n = \frac{2x^2}{(1-x)^3}$$

Step 4: Substitute $$x = \frac{1}{7}$$

Substitute $$x = \frac{1}{7}$$ into the formula:

$$\sum_{n=2}^{\infty} n(n-1) \left(\frac{1}{7}\right)^n = \frac{2\left(\frac{1}{7}\right)^2}{\left(1 - \frac{1}{7}\right)^3}$$  $$\sum_{n=2}^{\infty} n(n-1) \left(\frac{1}{7}\right)^n = \frac{\frac{2}{49}}{\left(\frac{6}{7}\right)^3} = \frac{\frac{2}{49}}{\frac{216}{343}} = \frac{2}{49} \times \frac{343}{216}$$

Since $$\frac{343}{49} = 7$$ and $$\frac{2}{216} = \frac{1}{108}$$:

$$\sum_{n=2}^{\infty} n(n-1) \left(\frac{1}{7}\right)^n = \frac{7}{108}$$

Step 5: Calculate final value of $$S$$

Remember that $$S = \frac{1}{2} \sum_{n=2}^{\infty} n(n-1) \left(\frac{1}{7}\right)^n$$:

$$S = \frac{1}{2} \times \frac{7}{108} = \frac{7}{216}$$

Answer:

$$\frac{7}{216}$$