A computer generates even integers half of the time and another computer generates even integers a third of the time. If $a_i$ and $b_i$ are the integers generated by the computers, respectively, at time $i$, what is the probability that $a_1b_1 +a_2b_2 +\cdots + a_kb_k$ is an even integer.
1. **Define the problem and initial conditions:**
   - Let \( p_k \) be the probability that \( \sum_{i=1}^{k} a_i b_i \) is even.
   - The probability that \( a_i \) is even is \( \frac{1}{2} \).
   - The probability that \( b_i \) is even is \( \frac{1}{3} \).

2. **Calculate the probability that \( a_i b_i \) is even:**
   - \( a_i b_i \) is even if either \( a_i \) or \( b_i \) (or both) are even.
   - The probability that \( a_i \) is even and \( b_i \) is even is \( \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \).
   - The probability that \( a_i \) is even and \( b_i \) is odd is \( \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \).
   - The probability that \( a_i \) is odd and \( b_i \) is even is \( \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \).
   - Therefore, the total probability that \( a_i b_i \) is even is:
     \[
     \frac{1}{6} + \frac{1}{3} + \frac{1}{6} = \frac{2}{3}
     \]

3. **Calculate the probability that \( a_i b_i \) is odd:**
   - The probability that \( a_i \) is odd and \( b_i \) is odd is \( \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \).

4. **Formulate the recursive relation for \( p_{k+1} \):**
   - The probability that \( \sum_{i=1}^{k+1} a_i b_i \) is even can be derived from the previous step:
     \[
     p_{k+1} = p_k \cdot \frac{2}{3} + (1 - p_k) \cdot \frac{1}{3}
     \]
   - Simplify the expression:
     \[
     p_{k+1} = \frac{2}{3} p_k + \frac{1}{3} (1 - p_k) = \frac{2}{3} p_k + \frac{1}{3} - \frac{1}{3} p_k = \frac{1}{3} p_k + \frac{1}{3}
     \]

5. **Solve the recursive relation:**
   - Let \( q_k = p_k - \frac{1}{2} \). Then:
     \[
     q_{k+1} = \frac{1}{3} q_k
     \]
   - This is a geometric sequence with \( q_1 = p_1 - \frac{1}{2} \).
   - Since \( p_1 = \frac{2}{3} \) (the probability that \( a_1 b_1 \) is even):
     \[
     q_1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}
     \]
   - Therefore:
     \[
     q_k = \left( \frac{1}{3} \right)^{k-1} q_1 = \left( \frac{1}{3} \right)^{k-1} \cdot \frac{1}{6}
     \]
   - Converting back to \( p_k \):
     \[
     p_k = q_k + \frac{1}{2} = \left( \frac{1}{3} \right)^{k-1} \cdot \frac{1}{6} + \frac{1}{2}
     \]

6. **Verify the solution for \( k = 0 \):**
   - For \( k = 0 \):
     \[
     p_0 = \frac{1}{6} \left( \frac{1}{3} \right)^{-1} + \frac{1}{2} = \frac{1}{6} \cdot 3 + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1
     \]

The final answer is \( \boxed{ p_k = \frac{1}{6} \left( \frac{1}{3} \right)^{k-1} + \frac{1}{2} } \)