To find the sum of the first 6 terms of the given infinite series [tex]\(1 - 2 + 4 - 8 + \cdots\)[/tex], let's proceed with the following steps:
1. Identify the first term [tex]\(a\)[/tex] and the common ratio [tex]\(r\)[/tex]:
- The first term [tex]\(a\)[/tex] is 1.
- The common ratio [tex]\(r\)[/tex] can be determined by dividing the second term by the first term: [tex]\(-2 / 1 = -2\)[/tex].
2. Write down the formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series:
For a geometric series, the sum of the first [tex]\(n\)[/tex] terms [tex]\(S_n\)[/tex] is given by:
[tex]\[
S_n = a \frac{1 - r^n}{1 - r}
\][/tex]
where:
- [tex]\(a\)[/tex] is the first term,
- [tex]\(r\)[/tex] is the common ratio,
- [tex]\(n\)[/tex] is the number of terms.
3. Substitute the known values into the formula:
- Here, the first term [tex]\(a = 1\)[/tex],
- The common ratio [tex]\(r = -2\)[/tex],
- And the number of terms [tex]\(n = 6\)[/tex].
So the sum of the first 6 terms [tex]\(S_6\)[/tex] is:
[tex]\[
S_6 = 1 \frac{1 - (-2)^6}{1 - (-2)}
\][/tex]
4. Simplify the expression step-by-step:
- Calculate [tex]\((-2)^6\)[/tex]: [tex]\((-2)^6 = 64\)[/tex],
- Next, substitute that back into the sum formula:
[tex]\[
S_6 = 1 \frac{1 - 64}{1 + 2}
\][/tex]
- Simplify the numerator:
[tex]\[
1 - 64 = -63
\][/tex]
- And the denominator:
[tex]\[
1 + 2 = 3
\][/tex]
- Therefore:
[tex]\[
S_6 = \frac{-63}{3} = -21
\][/tex]
5. State the final answer:
The sum of the first 6 terms of the series [tex]\(1 - 2 + 4 - 8 + \cdots\)[/tex] is [tex]\(-21\)[/tex].
