`sum_(k=1)^(oo)sum_(r=1)^(k)1/(4^(k))(""^(k)C_(r))` is equal to=________

Answer

To solve the given problem, we need to evaluate the double summation:

\[ \sum_{k=1}^{\infty} \sum_{r=1}^{k} \frac{1}{4^k} \binom{k}{r} \]

### Step 1: Rewrite the Inner Summation
The inner summation \(\sum_{r=1}^{k} \binom{k}{r}\) can be simplified. We know that:

\[ \sum_{r=0}^{k} \binom{k}{r} = 2^k \]

This includes the term for \(r=0\), which is \(\binom{k}{0} = 1\). Therefore, we can write:

\[ \sum_{r=1}^{k} \binom{k}{r} = 2^k - 1 \]

### Step 2: Substitute Back into the Outer Summation
Now we substitute this result back into the outer summation:

\[ \sum_{k=1}^{\infty} \frac{1}{4^k} (2^k - 1) \]

This can be separated into two summations:

\[ \sum_{k=1}^{\infty} \frac{2^k}{4^k} - \sum_{k=1}^{\infty} \frac{1}{4^k} \]

### Step 3: Simplify Each Summation
The first summation simplifies as follows:

\[ \sum_{k=1}^{\infty} \frac{2^k}{4^k} = \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^k \]

This is a geometric series with first term \(a = \frac{1}{2}\) and common ratio \(r = \frac{1}{2}\):

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

The second summation is also a geometric series:

\[ \sum_{k=1}^{\infty} \frac{1}{4^k} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

### Step 4: Combine the Results
Now we can combine the results from the two summations:

\[ 1 - \frac{1}{3} = \frac{2}{3} \]

### Final Answer
Thus, the value of the original double summation is:

\[ \sum_{k=1}^{\infty} \sum_{r=1}^{k} \frac{1}{4^k} \binom{k}{r} = \frac{2}{3} \]