IDNLearn.com helps you find the answers you need quickly and efficiently. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
Sagot :
Certainly! We are given the sequence defined by the formula:
[tex]\[ a_n = \frac{(-1)^n}{2n - 1} \][/tex]
Let's find the value of [tex]\(a_n\)[/tex] for a specific [tex]\(n\)[/tex]. We can proceed step-by-step:
1. Identify the value of [tex]\(n\)[/tex]:
- Let's choose [tex]\(n = 1\)[/tex].
2. Substitute [tex]\(n = 1\)[/tex] into the formula:
[tex]\[ a_1 = \frac{(-1)^1}{2 \cdot 1 - 1} \][/tex]
3. Evaluate the numerator:
- Calculate [tex]\((-1)^1\)[/tex]:
[tex]\[ (-1)^1 = -1 \][/tex]
4. Evaluate the denominator:
- Calculate [tex]\(2 \cdot 1 - 1\)[/tex]:
[tex]\[ 2 \cdot 1 - 1 = 2 - 1 = 1 \][/tex]
5. Form the fraction:
- Now, we have:
[tex]\[ a_1 = \frac{-1}{1} = -1 \][/tex]
Thus, the value of [tex]\( a_1 \)[/tex] is:
[tex]\[ a_1 = -1.0 \][/tex]
This completes our step-by-step solution for finding [tex]\(a_n\)[/tex] at [tex]\(n = 1\)[/tex].
[tex]\[ a_n = \frac{(-1)^n}{2n - 1} \][/tex]
Let's find the value of [tex]\(a_n\)[/tex] for a specific [tex]\(n\)[/tex]. We can proceed step-by-step:
1. Identify the value of [tex]\(n\)[/tex]:
- Let's choose [tex]\(n = 1\)[/tex].
2. Substitute [tex]\(n = 1\)[/tex] into the formula:
[tex]\[ a_1 = \frac{(-1)^1}{2 \cdot 1 - 1} \][/tex]
3. Evaluate the numerator:
- Calculate [tex]\((-1)^1\)[/tex]:
[tex]\[ (-1)^1 = -1 \][/tex]
4. Evaluate the denominator:
- Calculate [tex]\(2 \cdot 1 - 1\)[/tex]:
[tex]\[ 2 \cdot 1 - 1 = 2 - 1 = 1 \][/tex]
5. Form the fraction:
- Now, we have:
[tex]\[ a_1 = \frac{-1}{1} = -1 \][/tex]
Thus, the value of [tex]\( a_1 \)[/tex] is:
[tex]\[ a_1 = -1.0 \][/tex]
This completes our step-by-step solution for finding [tex]\(a_n\)[/tex] at [tex]\(n = 1\)[/tex].
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.