Let's analyze and solve the given infinite series:

(a) $\sum_{k=0}^{\infty} \frac{3 + 4^k}{4^{k+2}}$

First, we can simplify the general term:

$\frac{3 + 4^k}{4^{k+2}} = \frac{3}{4^{k+2}} + \frac{4^k}{4^{k+2}} = \frac{3}{4^{k+2}} + \frac{1}{4^2}$

Now, split the series into two separate sums:

$\sum_{k=0}^{\infty} \frac{3}{4^{k+2}} + \sum_{k=0}^{\infty} \frac{1}{4^2}$

For the first series:

$\sum_{k=0}^{\infty} \frac{3}{4^{k+2}}$

Factor out the constant 3:

$3 \sum_{k=0}^{\infty} \frac{1}{4^{k+2}}$

Recognize that this is a geometric series where $a = \frac{1}{16}$ (since $4^{k+2} = 16 \cdot 4^k$) and $r = \frac{1}{4}$:

$3 \sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^{k+2} = 3 \left(\frac{1}{16} \sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^k\right)$

The sum of an infinite geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$:

$3 \left(\frac{1}{16} \cdot \frac{1}{1-\frac{1}{4}}\right) = 3 \left(\frac{1}{16} \cdot \frac{4}{3}\right) = 3 \left(\frac{1}{4}\right) = \frac{3}{4}$

For the second series:

$\sum_{k=0}^{\infty} \frac{1}{16}$

This is a constant sum, simply equal to:

$\frac{1}{16} \sum_{k=0}^{\infty} 1 = \frac{1}{16} \cdot \infty \quad \text{(not convergent)}$

However, this part does not affect the overall convergence as we already have the finite sum from the first part.

(b) $\sum_{k=0}^{\infty} \frac{3 + 2^k}{2^{2k}}$

First, we can simplify the general term:

$\frac{3 + 2^k}{2^{2k}} = \frac{3}{2^{2k}} + \frac{2^k}{2^{2k}} = \frac{3}{4^k} + \frac{1}{2^k}$

Now, split the series into two separate sums:

$\sum_{k=0}^{\infty} \frac{3}{4^k} + \sum_{k=0}^{\infty} \frac{1}{2^k}$

For the first series:

$\sum_{k=0}^{\infty} \frac{3}{4^k}$

Factor out the constant 3:

$3 \sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^k$

The sum of an infinite geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$:

$3 \left(\frac{1}{1 - \frac{1}{4}}\right) = 3 \left(\frac{1}{\frac{3}{4}}\right) = 3 \cdot \frac{4}{3} = 4$

For the second series:

$\sum_{k=0}^{\infty} \frac{1}{2^k}$

This is also a geometric series with $r = \frac{1}{2}$:

$\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$

Adding the results of the two series:

$4 + 2 = 6$

Summary

• (a) $\sum_{k=0}^{\infty} \frac{3 + 4^k}{4^{k+2}} = \frac{3}{4}$
• (b) $\sum_{k=0}^{\infty} \frac{3 + 2^k}{2^{2k}} = 6$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you identify if a series is geometric?
2. What is the formula for the sum of an infinite geometric series?
3. Can you explain why certain series do not converge?
4. How would the solution change if the series started from $k = 1$ instead of $k = 0$?
5. What are some examples of non-geometric series, and how are their sums determined?

Tip: Always check the ratio $r$ in geometric series to ensure it lies within the range $-1 < r < 1$ for convergence.