If \( \sum_{k=1}^{20} 3(2)^{1-k}=p \), write down \( \sum_{k=1}^{20} 24(2)^{-k} \) in terms of \( p \).

To rewrite \( \sum_{k=1}^{20} 24(2)^{-k} \) in terms of \( p \), we start by simplifying the goal expression. We have: \[ \sum_{k=1}^{20} 24(2)^{-k} = 24 \sum_{k=1}^{20} (2)^{-k} \] Next, we need to recognize how to relate this sum to the one provided in \( p \). We note that: \[ \sum_{k=1}^{20} (2)^{-k} = \sum_{k=1}^{20} \frac{1}{(2)^k} \] This sum is a geometric series with the first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{2} \). The number of terms is \( n = 20 \). The formula for the sum of the first \( n \) terms of a geometric series is: \[ S_n = a \frac{1-r^n}{1-r} \] Substituting \( a \), \( r \), and \( n \): \[ \sum_{k=1}^{20} (2)^{-k} = \frac{\frac{1}{2}(1 - (\frac{1}{2})^{20})}{1 - \frac{1}{2}} = \frac{\frac{1}{2}(1 - \frac{1}{2^{20}})}{\frac{1}{2}} = 1 - \frac{1}{2^{20}} \] This means: \[ \sum_{k=1}^{20} 24(2)^{-k} = 24 \left( 1 - \frac{1}{2^{20}} \right) \] Next, let's analyze \( p = \sum_{k=1}^{20} 3(2)^{1-k} \): Rewriting it, we have: \[ p = 3 \sum_{k=1}^{20} (2)^{1-k} = 3(2) \sum_{k=1}^{20} (2)^{-k} = 6 \sum_{k=1}^{20} (2)^{-k} \] We can express \( \sum_{k=1}^{20} (2)^{-k} \) in terms of \( p \): \[ \sum_{k=1}^{20} (2)^{-k} = \frac{p}{6} \] Now substituting back into our expression for \( \sum_{k=1}^{20} 24(2)^{-k} \): \[ \sum_{k=1}^{20} 24(2)^{-k} = 24 \left( 1 - \frac{1}{2^{20}} \right) = 24 \left( 1 - \frac{1}{2^{20}} \right) \] Putting it all together, we get: \[ \sum_{k=1}^{20} 24(2)^{-k} = 24 \left( 1 - \frac{1}{2^{20}} \right) = 4p - 24\frac{1}{2^{20}} \] Thus, expressing \( \sum_{k=1}^{20} 24(2)^{-k} \) in terms of \( p \) gives: \[ \sum_{k=1}^{20} 24(2)^{-k} = 4p - \frac{3}{2^{19}} \] In conclusion: \[ \sum_{k=1}^{20} 24(2)^{-k} = 4p - \frac{3}{2^{19}} \]