Let's solve the given mathematical series step by step.
We are asked to find the approximate value of the series [tex]\(\sum_{n=1}^8 2.5(0.5)^{n-1}\)[/tex] and round the answer to the nearest hundredth.
This is a geometric series where the first term [tex]\(a = 2.5\)[/tex] and the common ratio [tex]\(r = 0.5\)[/tex]. In a geometric series, each term is obtained by multiplying the previous term by a constant ratio [tex]\(r\)[/tex].
The general formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series [tex]\(S_n\)[/tex] is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
In this series:
- [tex]\(a = 2.5\)[/tex]
- [tex]\(r = 0.5\)[/tex]
- [tex]\(n = 8\)[/tex]
Plugging these values into the formula gives:
[tex]\[ S_8 = 2.5 \frac{1 - (0.5)^8}{1 - 0.5} \][/tex]
First, we calculate [tex]\((0.5)^8\)[/tex]:
[tex]\[ (0.5)^8 = 0.00390625 \][/tex]
Next, we subtract this value from 1:
[tex]\[ 1 - 0.00390625 = 0.99609375 \][/tex]
Now, we divide this result by [tex]\((1 - 0.5 = 0.5)\)[/tex]:
[tex]\[ \frac{0.99609375}{0.5} = 1.9921875 \][/tex]
Finally, multiply by the first term [tex]\(a = 2.5\)[/tex]:
[tex]\[ S_8 = 2.5 \times 1.9921875 = 4.98046875 \][/tex]
Rounding this result to the nearest hundredth:
[tex]\[ 4.98046875 \approx 4.98 \][/tex]
Thus, the approximate value of the series [tex]\(\sum_{n=1}^8 2.5(0.5)^{n-1}\)[/tex] is [tex]\(\boxed{4.98}\)[/tex].
