IDNLearn.com makes it easy to find precise answers to your specific questions. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
Sagot :
Certainly! Let's evaluate the infinite series:
[tex]\[ \sum_{n=1}^{\infty} \frac{9^n}{n (-2)^{n+1}}. \][/tex]
1. Identify the General Term:
Let's denote the general term of the series by [tex]\( a_n \)[/tex]:
[tex]\[ a_n = \frac{9^n}{n (-2)^{n+1}}. \][/tex]
2. Simplify the General Term:
We notice that [tex]\( (-2)^{n+1} = (-2) \cdot (-2)^n \)[/tex]:
[tex]\[ a_n = \frac{9^n}{n \cdot (-2) \cdot (-2)^n} = \frac{9^n}{-2 \cdot n \cdot (-2)^n}. \][/tex]
3. Combine the Powers in the Denominator:
Combine [tex]\( (-2) \cdot (-2)^n \)[/tex]:
[tex]\[ (-2) \cdot (-2)^n = (-1)^{n+1} \cdot 2^n. \][/tex]
Therefore,
[tex]\[ a_n = \frac{9^n}{n \cdot (-1)^{n+1} \cdot 2^n}. \][/tex]
4. Write the Combined General Term:
Bring [tex]\( (-1)^{n+1} \)[/tex] to the numerator:
[tex]\[ a_n = \frac{9^n}{n \cdot 2^n} \cdot \frac{1}{(-1)^{n+1}} = \frac{9^n}{n \cdot 2^n} \cdot (-1)^{-(n+1)}. \][/tex]
5. Simplify the Exponent:
We know that [tex]\( (-1)^{-(n+1)} = (-1)^{-n-1} \)[/tex], so:
[tex]\[ a_n = (-1)^{-n-1} \cdot \frac{9^n}{n \cdot 2^n}. \][/tex]
6. Combine the Powers of the Numerator and Denominator:
Combine the powers of 9 / 2:
[tex]\[ a_n = \left(\frac{9}{2}\right)^n \cdot (-1)^{-n-1} \cdot \frac{1}{n}. \][/tex]
7. Express the General Term in a Simplified Form:
The simplified general term is:
[tex]\[ a_n = \frac{9^n}{n \cdot (-2)^{n+1}}. \][/tex]
Now, we can write the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{9^n}{n (-2)^{n+1}} = \sum_{n=1}^{\infty} \frac{(9 / 2)^n}{n} \cdot (-1)^{-(n+1)}. \][/tex]
Final Summation Representation:
The sum of the given series can be expressed as:
[tex]\[ \boxed{\sum_{n=1}^{\infty} \frac{9^n}{n(-2)^{n+1}} = \sum_{n=1}^{\infty} \frac{(9 / 2)^n}{n} \cdot (-1)^{-(n+1)}}. \][/tex]
This is the most simplified form of the given series representation.
[tex]\[ \sum_{n=1}^{\infty} \frac{9^n}{n (-2)^{n+1}}. \][/tex]
1. Identify the General Term:
Let's denote the general term of the series by [tex]\( a_n \)[/tex]:
[tex]\[ a_n = \frac{9^n}{n (-2)^{n+1}}. \][/tex]
2. Simplify the General Term:
We notice that [tex]\( (-2)^{n+1} = (-2) \cdot (-2)^n \)[/tex]:
[tex]\[ a_n = \frac{9^n}{n \cdot (-2) \cdot (-2)^n} = \frac{9^n}{-2 \cdot n \cdot (-2)^n}. \][/tex]
3. Combine the Powers in the Denominator:
Combine [tex]\( (-2) \cdot (-2)^n \)[/tex]:
[tex]\[ (-2) \cdot (-2)^n = (-1)^{n+1} \cdot 2^n. \][/tex]
Therefore,
[tex]\[ a_n = \frac{9^n}{n \cdot (-1)^{n+1} \cdot 2^n}. \][/tex]
4. Write the Combined General Term:
Bring [tex]\( (-1)^{n+1} \)[/tex] to the numerator:
[tex]\[ a_n = \frac{9^n}{n \cdot 2^n} \cdot \frac{1}{(-1)^{n+1}} = \frac{9^n}{n \cdot 2^n} \cdot (-1)^{-(n+1)}. \][/tex]
5. Simplify the Exponent:
We know that [tex]\( (-1)^{-(n+1)} = (-1)^{-n-1} \)[/tex], so:
[tex]\[ a_n = (-1)^{-n-1} \cdot \frac{9^n}{n \cdot 2^n}. \][/tex]
6. Combine the Powers of the Numerator and Denominator:
Combine the powers of 9 / 2:
[tex]\[ a_n = \left(\frac{9}{2}\right)^n \cdot (-1)^{-n-1} \cdot \frac{1}{n}. \][/tex]
7. Express the General Term in a Simplified Form:
The simplified general term is:
[tex]\[ a_n = \frac{9^n}{n \cdot (-2)^{n+1}}. \][/tex]
Now, we can write the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{9^n}{n (-2)^{n+1}} = \sum_{n=1}^{\infty} \frac{(9 / 2)^n}{n} \cdot (-1)^{-(n+1)}. \][/tex]
Final Summation Representation:
The sum of the given series can be expressed as:
[tex]\[ \boxed{\sum_{n=1}^{\infty} \frac{9^n}{n(-2)^{n+1}} = \sum_{n=1}^{\infty} \frac{(9 / 2)^n}{n} \cdot (-1)^{-(n+1)}}. \][/tex]
This is the most simplified form of the given series representation.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.