Question

Compute
1 + \frac{3}{4} + \frac{5}{4^2} + \frac{7}{4^3}

Answer 1

To compute the sum

\[
S = 1 + \frac{3}{4} + \frac{5}{4^2} + \frac{7}{4^3},
\]

we first rewrite the terms systematically.

The first term is

\[
1 = \frac{1 \cdot 4^3}{4^3} = \frac{4^3}{4^3} = \frac{64}{64}.
\]

The second term is

\[
\frac{3}{4} = \frac{3 \cdot 4^2}{4^3} = \frac{12}{64}.
\]

The third term is

\[
\frac{5}{4^2} = \frac{5 \cdot 4}{4^3} = \frac{20}{64}.
\]

The fourth term is

\[
\frac{7}{4^3} = \frac{7}{64}.
\]

Now we can find a common denominator:

\[
S = \frac{64}{64} + \frac{12}{64} + \frac{20}{64} + \frac{7}{64} = \frac{64 + 12 + 20 + 7}{64}.
\]

Calculating the numerator:

\[
64 + 12 = 76,
\]
\[
76 + 20 = 96,
\]
\[
96 + 7 = 103.
\]

Thus, we find

\[
S = \frac{103}{64}.
\]

Thus, the final answer is

\[
\boxed{\frac{103}{64}}.
\]