Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine which summation represents the total number of shaded triangles in the first 15 figures of the Sierpinski triangle, let's analyze each of the provided summations one by one:
1. [tex]\(\sum_{n=1}^{15} 1(3)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( 1 (3)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( 1 (3)^{1-1} = 1 (3)^0 = 1 \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( 1 (3)^{2-1} = 1 (3)^1 = 3 \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( 1 (3)^{3-1} = 1 (3)^2 = 9 \)[/tex]
- And so on...
- This pattern matches the one given (1, 3, 9...).
Hence, the total number of shaded triangles in the first 15 figures would be:
[tex]\[ 1 (3)^{15-1} = 4782969 \][/tex]
2. [tex]\(\sum_{n=1}^{15} 3(1)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( 3 (1)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( 3 (1)^0 = 3 \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( 3 (1)^1 = 3 \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( 3 (1)^2 = 3 \)[/tex]
- And so on...
- The result for each term in this series is always [tex]\( 3 \)[/tex].
Hence, the total for the first 15 figures is [tex]\( 3 \times 15 = 45 \)[/tex]. However, considering only the last term as given:
[tex]\[ 3 (1)^{15-1} = 3 \][/tex]
3. [tex]\(\sum_{n=1}^{15} 1\left(\frac{1}{3}\right)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( 1 \left( \frac{1}{3} \right)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( 1 \left( \frac{1}{3} \right)^0 = 1 \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( 1 \left( \frac{1}{3} \right)^1 = \frac{1}{3} \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( 1 \left( \frac{1}{3} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \)[/tex]
- The terms get smaller (fractions less than 1).
Considering the last term for [tex]\( n = 15 \)[/tex]:
[tex]\[ 1\left(\frac{1}{3}\right)^{15-1} = 2.090751581287688 \times 10^{-7} \][/tex]
4. [tex]\(\sum_{n=1}^{15} \frac{1}{3}(1)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( \frac{1}{3} (1)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( \frac{1}{3} (1)^0 = \frac{1}{3} \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( \frac{1}{3} (1)^1 = \frac{1}{3} \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( \frac{1}{3} (1)^2 = \frac{1}{3} \)[/tex]
- The value remains [tex]\( \frac{1}{3} \)[/tex] for each term.
Hence, for the last term at [tex]\( n = 15 \)[/tex]:
[tex]\[ \frac{1}{3} (1)^{15-1} = \frac{1}{3} = 0.3333333333333333 \][/tex]
From analyzing all the given options, it is evident that the correct summation representing the total number of shaded triangles in the first 15 figures is:
[tex]\[ \sum_{n=1}^{15} 1(3)^{n-1} \][/tex]
Therefore, we select:
\[
\sum_{n=1}^{15} 1(3)^{n-1}.
1. [tex]\(\sum_{n=1}^{15} 1(3)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( 1 (3)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( 1 (3)^{1-1} = 1 (3)^0 = 1 \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( 1 (3)^{2-1} = 1 (3)^1 = 3 \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( 1 (3)^{3-1} = 1 (3)^2 = 9 \)[/tex]
- And so on...
- This pattern matches the one given (1, 3, 9...).
Hence, the total number of shaded triangles in the first 15 figures would be:
[tex]\[ 1 (3)^{15-1} = 4782969 \][/tex]
2. [tex]\(\sum_{n=1}^{15} 3(1)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( 3 (1)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( 3 (1)^0 = 3 \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( 3 (1)^1 = 3 \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( 3 (1)^2 = 3 \)[/tex]
- And so on...
- The result for each term in this series is always [tex]\( 3 \)[/tex].
Hence, the total for the first 15 figures is [tex]\( 3 \times 15 = 45 \)[/tex]. However, considering only the last term as given:
[tex]\[ 3 (1)^{15-1} = 3 \][/tex]
3. [tex]\(\sum_{n=1}^{15} 1\left(\frac{1}{3}\right)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( 1 \left( \frac{1}{3} \right)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( 1 \left( \frac{1}{3} \right)^0 = 1 \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( 1 \left( \frac{1}{3} \right)^1 = \frac{1}{3} \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( 1 \left( \frac{1}{3} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \)[/tex]
- The terms get smaller (fractions less than 1).
Considering the last term for [tex]\( n = 15 \)[/tex]:
[tex]\[ 1\left(\frac{1}{3}\right)^{15-1} = 2.090751581287688 \times 10^{-7} \][/tex]
4. [tex]\(\sum_{n=1}^{15} \frac{1}{3}(1)^{n-1}\)[/tex]
- This summation considers the expression [tex]\( \frac{1}{3} (1)^{n-1} \)[/tex].
- When [tex]\( n = 1 \)[/tex]: [tex]\( \frac{1}{3} (1)^0 = \frac{1}{3} \)[/tex]
- When [tex]\( n = 2 \)[/tex]: [tex]\( \frac{1}{3} (1)^1 = \frac{1}{3} \)[/tex]
- When [tex]\( n = 3 \)[/tex]: [tex]\( \frac{1}{3} (1)^2 = \frac{1}{3} \)[/tex]
- The value remains [tex]\( \frac{1}{3} \)[/tex] for each term.
Hence, for the last term at [tex]\( n = 15 \)[/tex]:
[tex]\[ \frac{1}{3} (1)^{15-1} = \frac{1}{3} = 0.3333333333333333 \][/tex]
From analyzing all the given options, it is evident that the correct summation representing the total number of shaded triangles in the first 15 figures is:
[tex]\[ \sum_{n=1}^{15} 1(3)^{n-1} \][/tex]
Therefore, we select:
\[
\sum_{n=1}^{15} 1(3)^{n-1}.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
