Question: Which summation formula is equivalent to this series? \[ \sum_{n=0}^{3} \left( -\frac{1}{2}…

Which summation formula is equivalent to this series?

\[ \sum_{n=0}^{3} \left( -\frac{1}{2} \right)^n (4) \]

Options:

1. \(\sum_{n=-4}^{1} (2)^{-n}\)
2. \(\sum_{n=2}^{5} (-2)^n\)
3. \(\sum_{n=0}^{3} (-2)^{2n}\)
4. \(\sum_{n=-4}^{1} \left( \frac{1}{4} \right) (-2)^{-n}\)

Solution

To determine which summation formula is equivalent to the series: \[ \sum_{n=0}^{3} \left(\frac{1}{2}\right)^n (4) \] let’s simplify it. First, evaluate the expression inside the summation: \[ \left(\frac{1}{2}\right)^n \cdot 4 = 4 \cdot \left(\frac{1}{2}\right)^n = 4 \cdot 2^{-n} = (2^2) \cdot 2^{-n} = 2^{2-n} \] Now, the original series becomes: \[ \sum_{n=0}^{3} 2^{2-n} \] Let’s check which of the options matches this expression: 1. \(\sum_{n=-4}^{-1} (2)^{-n}\) 2. \(\sum_{n=2}^{5} (-2)^n\) 3. \(\sum_{n=0}^{3} (-2)^{2n}\) 4. \(\sum_{n=-4}^{-1} \left(\frac{1}{4}\right) (-2)^{-n}\) Option 1 matches our expression: \[ 2^{2-n} = 2^{-n+2} = (2^2) \cdot (2)^{-n} \] So the equivalent summation formula is: \[ \sum_{n=-4}^{-1} (2)^{-n} \]