To solve the given problem, we need to determine the sum of the series [tex]\(\sum_{k=1}^7 729\left(3^{1-k}\right)\)[/tex].
1. Understanding the Series:
The given series is:
[tex]\[
S = \sum_{k=1}^7 729 \left(3^{1-k}\right)
\][/tex]
This means:
[tex]\[
S = 729 \left(3^{1-1}\right) + 729 \left(3^{1-2}\right) + 729 \left(3^{1-3}\right) + \cdots + 729 \left(3^{1-7}\right)
\][/tex]
2. Simplifying the Terms:
Each term inside the series can be simplified. When [tex]\( k = 1 \)[/tex]:
[tex]\[
729 \left(3^{1-1}\right) = 729 \left(3^0\right) = 729 \cdot 1 = 729
\][/tex]
When [tex]\( k = 2 \)[/tex]:
[tex]\[
729 \left(3^{1-2}\right) = 729 \left(3^{-1}\right) = 729 \cdot \frac{1}{3} = 243
\][/tex]
When [tex]\( k = 3 \)[/tex]:
[tex]\[
729 \left(3^{1-3}\right) = 729 \left(3^{-2}\right) = 729 \cdot \left(\frac{1}{3^2}\right) = 729 \cdot \frac{1}{9} = 81
\][/tex]
When [tex]\( k = 4 \)[/tex]:
[tex]\[
729 \left(3^{1-4}\right) = 729 \left(3^{-3}\right) = 729 \cdot \left(\frac{1}{3^3}\right) = 729 \cdot \frac{1}{27} = 27
\][/tex]
When [tex]\( k = 5 \)[/tex]:
[tex]\[
729 \left(3^{1-5}\right) = 729 \left(3^{-4}\right) = 729 \cdot \left(\frac{1}{3^4}\right) = 729 \cdot \frac{1}{81} = 9
\][/tex]
When [tex]\( k = 6 \)[/tex]:
[tex]\[
729 \left(3^{1-6}\right) = 729 \left(3^{-5}\right) = 729 \cdot \left(\frac{1}{3^5}\right) = 729 \cdot \frac{1}{243} = 3
\][/tex]
When [tex]\( k = 7 \)[/tex]:
[tex]\[
729 \left(3^{1-7}\right) = 729 \left(3^{-6}\right) = 729 \cdot \left(\frac{1}{3^6}\right) = 729 \cdot \frac{1}{729} = 1
\][/tex]
3. Summing the Series:
Now, we'll add up these terms:
[tex]\[
729 + 243 + 81 + 27 + 9 + 3 + 1
\][/tex]
4. Calculating the Sum:
[tex]\[
729 + 243 = 972
\][/tex]
[tex]\[
972 + 81 = 1053
\][/tex]
[tex]\[
1053 + 27 = 1080
\][/tex]
[tex]\[
1080 + 9 = 1089
\][/tex]
[tex]\[
1089 + 3 = 1092
\][/tex]
[tex]\[
1092 + 1 = 1093
\][/tex]
Therefore, the sum is [tex]\(1093\)[/tex].
Answer: B. 1093
