Question

Which series is represented by:
Σ^5 n=1 64 (1/4)^n-1

A. 16,384 + 4096 + 1024 + 256 + 6,4
B. 1/4 + 1 + 4 + 16 + 64
C. 64 + 256 + 1024 + 4096 + 16,384
D. 64 + 16 + 4 + 1 + 1

Answer 1

To evaluate the series represented by \(\sum_{n=1}^5 64 \left(\frac{1}{4}\right)^{n-1}\), we can simplify this into a more manageable form.

This expression can be expanded as follows:

\[
\sum_{n=1}^5 64 \left(\frac{1}{4}\right)^{n-1} = 64 \left(\frac{1}{4}\right)^0 + 64 \left(\frac{1}{4}\right)^1 + 64 \left(\frac{1}{4}\right)^2 + 64 \left(\frac{1}{4}\right)^3 + 64 \left(\frac{1}{4}\right)^4
\]

Calculating each term gives us:

1. For \(n=1\):
\[
64 \left(\frac{1}{4}\right)^{0} = 64 \times 1 = 64
\]

2. For \(n=2\):
\[
64 \left(\frac{1}{4}\right)^{1} = 64 \times \frac{1}{4} = 16
\]

3. For \(n=3\):
\[
64 \left(\frac{1}{4}\right)^{2} = 64 \times \left(\frac{1}{16}\right) = 4
\]

4. For \(n=4\):
\[
64 \left(\frac{1}{4}\right)^{3} = 64 \times \left(\frac{1}{64}\right) = 1
\]

5. For \(n=5\):
\[
64 \left(\frac{1}{4}\right)^{4} = 64 \times \left(\frac{1}{256}\right) = \frac{64}{256} = \frac{1}{4}
\]

Adding all these terms together:

\[
64 + 16 + 4 + 1 + \frac{1}{4}
\]

If we combine them:

- The sum of \(64 + 16 + 4 + 1 = 85\).
- Adding \(\frac{1}{4}\) gives \(85 + \frac{1}{4} = 85.25\).

Now we will evaluate the options presented:

A. \(16,384 + 4096 + 1024 + 256 + 6.4\) - This does not fit.

B. \(1/4 + 1 + 4 + 16 + 64\) - This gives \(64 + 16 + 4 + 1 + 0.25\) which sums to \(85.25\), matching our result.

C. \(64 + 256 + 1024 + 4096 + 16,384\) - This sum is much larger than our result.

D. \(64 + 16 + 4 + 1 + 1\) - This sums to \(86\) - again does not match.

Therefore, the correct answer to the series is:

**B. \(1/4 + 1 + 4 + 16 + 64\)**.