The problem asks us to evaluate the infinite sum $\sum_{k=1}^{\infty} \frac{4^{k+1}}{7^{k-1}}$.

1. Problem Description

The problem asks us to evaluate the infinite sum $\sum_{k=1}^{\infty} \frac{4^{k+1}}{7^{k-1}}$.

2. Solution Steps

We can rewrite the given sum as follows:

$\sum_{k=1}^{\infty} \frac{4^{k+1}}{7^{k-1}} = \sum_{k=1}^{\infty} \frac{4^k \cdot 4}{7^k \cdot 7^{-1}} = \sum_{k=1}^{\infty} \frac{4 \cdot 7 \cdot 4^k}{7^k} = 28 \sum_{k=1}^{\infty} \left(\frac{4}{7}\right)^k$.

The sum $\sum_{k=1}^{\infty} \left(\frac{4}{7}\right)^k$ is a geometric series with first term $a = \frac{4}{7}$ and common ratio $r = \frac{4}{7}$.

Since $|r| = \left|\frac{4}{7}\right| < 1$, the geometric series converges, and its sum is given by

$\sum_{k=1}^{\infty} ar^{k-1} = \frac{a}{1-r}$.

In our case, we have $\sum_{k=1}^{\infty} \left(\frac{4}{7}\right)^k = \sum_{k=1}^{\infty} \frac{4}{7} \cdot \left(\frac{4}{7}\right)^{k-1}$.

So the sum is $\frac{\frac{4}{7}}{1-\frac{4}{7}} = \frac{\frac{4}{7}}{\frac{3}{7}} = \frac{4}{3}$.

Then, we have

$28 \sum_{k=1}^{\infty} \left(\frac{4}{7}\right)^k = 28 \cdot \frac{4}{3} = \frac{112}{3}$.

3. Final Answer

$\frac{112}{3}$