Question
What is the sum of the geometric series LaTeX: \sum_{k=1}^{10}{6(2)^k}?
Answers
GPT-4o mini
To find the sum of the geometric series given by
\[
\sum_{k=1}^{10}6(2)^k,
\]
we can first rewrite this sum in a more conventional form. The series can be expressed as:
\[
6 \sum_{k=1}^{10}(2^k).
\]
Now, we need to identify the elements of the geometric series. The first term \(a\) of the series when \(k=1\) is:
\[
a = 2^1 = 2.
\]
The common ratio \(r\) is:
\[
r = 2.
\]
The number of terms \(n\) in the series is 10 (from \(k=1\) to \(k=10\)). We can use the formula for the sum of the first \(n\) terms of a geometric series:
\[
S_n = a \frac{r^n - 1}{r - 1}.
\]
Substituting the values of \(a\), \(r\), and \(n\):
\[
S_{10} = 2 \frac{2^{10} - 1}{2 - 1} = 2 \frac{1024 - 1}{1} = 2 \times 1023 = 2046.
\]
Now, we multiply by the factor 6 we factored out earlier:
\[
6 \cdot S_{10} = 6 \cdot 2046 = 12276.
\]
Thus, the sum of the geometric series is
\[
\boxed{12276}.
\]
\[
\sum_{k=1}^{10}6(2)^k,
\]
we can first rewrite this sum in a more conventional form. The series can be expressed as:
\[
6 \sum_{k=1}^{10}(2^k).
\]
Now, we need to identify the elements of the geometric series. The first term \(a\) of the series when \(k=1\) is:
\[
a = 2^1 = 2.
\]
The common ratio \(r\) is:
\[
r = 2.
\]
The number of terms \(n\) in the series is 10 (from \(k=1\) to \(k=10\)). We can use the formula for the sum of the first \(n\) terms of a geometric series:
\[
S_n = a \frac{r^n - 1}{r - 1}.
\]
Substituting the values of \(a\), \(r\), and \(n\):
\[
S_{10} = 2 \frac{2^{10} - 1}{2 - 1} = 2 \frac{1024 - 1}{1} = 2 \times 1023 = 2046.
\]
Now, we multiply by the factor 6 we factored out earlier:
\[
6 \cdot S_{10} = 6 \cdot 2046 = 12276.
\]
Thus, the sum of the geometric series is
\[
\boxed{12276}.
\]