To find the sum of the geometric series [tex]\(\sum_{x=0}^{15} 2\left(\frac{1}{4}\right)^x\)[/tex] and round it to the nearest whole number, we need to follow these steps:
1. Identify the first term [tex]\(a\)[/tex] and the common ratio [tex]\(r\)[/tex]:
- The first term [tex]\(a\)[/tex] is the value of the series when [tex]\(x = 0\)[/tex].
[tex]\[ a = 2 \left(\frac{1}{4}\right)^0 = 2 \][/tex]
- The common ratio [tex]\(r\)[/tex] is the factor by which each term is multiplied to get the next term.
[tex]\[ r = \frac{1}{4} \][/tex]
2. Determine the number of terms [tex]\(n\)[/tex]:
- The series is given from [tex]\(x = 0\)[/tex] to [tex]\(x = 15\)[/tex], hence there are [tex]\(16\)[/tex] terms in total.
3. Use the formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series:
The formula for the sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of a geometric series is:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
- Here, [tex]\(a = 2\)[/tex], [tex]\(r = \frac{1}{4}\)[/tex], and [tex]\(n = 16\)[/tex].
4. Plug in the values into the formula:
[tex]\[ S_{16} = 2 \frac{1 - \left(\frac{1}{4}\right)^{16}}{1 - \frac{1}{4}} \][/tex]
5. Simplify the expression:
- Calculate [tex]\( \left(\frac{1}{4}\right)^{16} \)[/tex]:
[tex]\[ \left(\frac{1}{4}\right)^{16} \][/tex]
This term is very small, close to zero.
- Calculate the denominator [tex]\(1 - \frac{1}{4}\)[/tex]:
[tex]\[ 1 - \frac{1}{4} = \frac{3}{4} \][/tex]
- Simplify the fraction:
[tex]\[ S_{16} = 2 \frac{1 - \text{(something very small)}}{\frac{3}{4}} = 2 \frac{1}{\frac{3}{4}} = 2 \cdot \frac{4}{3} = \frac{8}{3} \][/tex]
6. Compute the sum:
- The value of the fraction:
[tex]\[ S_{16} = \frac{8}{3} \approx 2.666666666045785 \][/tex]
7. Round the result to the nearest whole number:
- [tex]\(2.666666666045785\)[/tex] rounded to the nearest whole number is [tex]\(3\)[/tex].
Therefore, the sum of the geometric series [tex]\(\sum_{x=0}^{15} 2\left(\frac{1}{4}\right)^x\)[/tex], rounded to the nearest whole number, is [tex]\(\boxed{3}\)[/tex].
