If $7=5+\dfrac{5+\alpha}{7}+\dfrac{5+2\alpha}{7^{2}}+\dfrac{5+3\alpha}{7^{3}}+\dotsb\infty$, then the value of $\alpha$ is:
Step-by-Step Solution
Key Concept: This is an arithmetico-geometric series $\sum_{n=0}^{\infty}\dfrac{5+n\alpha}{7^{n}}$. Split into $\sum 5/7^{n}$ (geometric) and $\alpha\sum n/7^{n}$ (well-known closed form $r/(1-r)^{2}$).
$$7=\sum_{n=0}^{\infty}\frac{5+n\alpha}{7^{n}}=5\sum_{n=0}^{\infty}\frac{1}{7^{n}}+\alpha\sum_{n=0}^{\infty}\frac{n}{7^{n}}.$$
With $r=1/7$: $\sum_{0}^{\infty}r^{n}=\dfrac{1}{1-r}=\dfrac{7}{6}$, and $\sum_{0}^{\infty}nr^{n}=\dfrac{r}{(1-r)^{2}}=\dfrac{1/7}{36/49}=\dfrac{7}{36}.$
$$7=5\cdot\frac{7}{6}+\alpha\cdot\frac{7}{36}=\frac{35}{6}+\frac{7\alpha}{36}.$$
$$7-\frac{35}{6}=\frac{7}{6}=\frac{7\alpha}{36}\Rightarrow \alpha=6.$$
Correct Answer: 2