Compute

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

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