Questions: Find the sum of the sequence. Sum from k=1 to 5 of (-1)^k 2^k Sum from k=1 to 5 of (-1)^k 2^k =

Solution Steps

To find the sum of the sequence \(\sum_{k=1}^{5}(-1)^{k} 2^{k}\), we need to evaluate the expression for each integer \(k\) from 1 to 5, apply the alternating sign \((-1)^{k}\), and then sum the results.

We need to evaluate the sum of the sequence given by

\[ \sum_{k=1}^{5}(-1)^{k} 2^{k}. \]

This means we will calculate the terms for \(k = 1\) to \(k = 5\):

• For \(k = 1\): \((-1)^{1} 2^{1} = -2\)
• For \(k = 2\): \((-1)^{2} 2^{2} = 4\)
• For \(k = 3\): \((-1)^{3} 2^{3} = -8\)
• For \(k = 4\): \((-1)^{4} 2^{4} = 16\)
• For \(k = 5\): \((-1)^{5} 2^{5} = -32\)

Now, we sum these evaluated terms:

\[ -2 + 4 - 8 + 16 - 32. \]

Calculating this step-by-step:

1. \(-2 + 4 = 2\)
2. \(2 - 8 = -6\)
3. \(-6 + 16 = 10\)
4. \(10 - 32 = -22\)

Final Answer