1. Write the given infinite series

We have

$$7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2\alpha)+\frac{1}{7^3}(5+3\alpha)+\cdots$$

This can be written as

$$7=5+\sum_{n=1}^{\infty}\frac{5+n\alpha}{7^n}$$

So,

$$7=5\left(1+\sum_{n=1}^{\infty}\frac{1}{7^n}\right)+\alpha\sum_{n=1}^{\infty}\frac{n}{7^n}$$

2. Evaluate the geometric series

We know

$$\sum_{n=1}^{\infty} r^n=\frac{r}{1-r}, \quad |r|<1$$

with $r=\frac17$.

Thus,

$$\sum_{n=1}^{\infty}\frac{1}{7^n}=\frac{\frac17}{1-\frac17}=\frac{1}{6}$$

Hence,

$$5\left(1+\sum_{n=1}^{\infty}\frac{1}{7^n}\right)=5\left(1+\frac16\right)=5\cdot\frac76=\frac{35}{6}$$

3. Evaluate the series involving $n$

We use the standard result

$$\sum_{n=1}^{\infty} nr^n=\frac{r}{(1-r)^2}, \quad |r|<1$$

Again with $r=\frac17$,

$$\sum_{n=1}^{\infty}\frac{n}{7^n}=\frac{\frac17}{\left(1-\frac17\right)^2} =\frac{\frac17}{\left(\frac67\right)^2} =\frac{1}{7}\cdot\frac{49}{36} =\frac{7}{36}$$

So the equation becomes

$$7=\frac{35}{6}+\alpha\cdot\frac{7}{36}$$

4. Solve for $\alpha$

Subtract $\frac{35}{6}$ from both sides:

$$7-\frac{35}{6}=\alpha\cdot\frac{7}{36}$$ $$\frac{42-35}{6}=\alpha\cdot\frac{7}{36}$$ $$\frac{7}{6}=\alpha\cdot\frac{7}{36}$$

Therefore,

$$\alpha=\frac{7}{6}\cdot\frac{36}{7}=6$$

5. Check with the options

The correct option is

$$\boxed{6}$$

which is Option D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.