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

Question

UpStudy AI · Accepted Answer

To express $$ \sum_{k=1}^{20} 24(2)^{-k} $$ in terms of $$ p $$, let's first analyze the given sum:

$$
p = \sum_{k=1}^{20} 3(2)^{1-k}
$$

We can rewrite $$ p $$ as:

$$
p = 3 \cdot 2 \cdot \sum_{k=1}^{20} 2^{-k} = 6 \cdot \sum_{k=1}^{20} 2^{-k}
$$

This simplifies to:

$$
\sum_{k=1}^{20} 2^{-k} = \frac{p}{6}
$$

Now, consider the sum we want to express in terms of $$ p $$:

$$
\sum_{k=1}^{20} 24(2)^{-k} = 24 \cdot \sum_{k=1}^{20} 2^{-k}
$$

Substituting the expression for $$ \sum_{k=1}^{20} 2^{-k} $$ from above:

$$
24 \cdot \sum_{k=1}^{20} 2^{-k} = 24 \cdot \frac{p}{6} = 4p
$$

**Answer:**  
It is equal to four times p. In symbols, the sum is 4 · p.
The sum $$ \sum_{k=1}^{20} 24(2)^{-k} $$ is equal to $$ 4p $$.

If , write down in terms of .