Answer :
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}.
