3. $sum_{n = 0}^{infty}(\frac{1}{6^{n}}-\frac{1}{7^{n}})=$\na. $\frac{1}{7}$\nb. $\frac{1}{30}$\nc…

Explanation:

Step1: Split the series

We know that $\sum_{n = 0}^{\infty}(\frac{1}{6^{n}}-\frac{1}{7^{n}})=\sum_{n = 0}^{\infty}\frac{1}{6^{n}}-\sum_{n = 0}^{\infty}\frac{1}{7^{n}}$.

Step2: Use the formula for the sum of an infinite - geometric series

The sum of an infinite geometric series $\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}$ for $|r|\lt1$. For $\sum_{n = 0}^{\infty}\frac{1}{6^{n}}$, where $r=\frac{1}{6}$, the sum is $\frac{1}{1-\frac{1}{6}}=\frac{1}{\frac{5}{6}}=\frac{6}{5}$. For $\sum_{n = 0}^{\infty}\frac{1}{7^{n}}$, where $r = \frac{1}{7}$, the sum is $\frac{1}{1-\frac{1}{7}}=\frac{1}{\frac{6}{7}}=\frac{7}{6}$.

Step3: Calculate the result

$\sum_{n = 0}^{\infty}\frac{1}{6^{n}}-\sum_{n = 0}^{\infty}\frac{1}{7^{n}}=\frac{6}{5}-\frac{7}{6}=\frac{36 - 35}{30}=\frac{1}{30}$.

Answer:

B. $\frac{1}{30}$