Question
Find the sum.
Summation from n equals 1 to 80 left parenthesis 2 n minus 8 right parenthesis
Summation from n equals 1 to 80 left parenthesis 2 n minus 8 right parenthesis
Answers
GPT-4o mini
To find the sum
\[
\sum_{n=1}^{80} (2n - 8),
\]
we can simplify the sum by distributing it:
\[
\sum_{n=1}^{80} (2n - 8) = \sum_{n=1}^{80} 2n - \sum_{n=1}^{80} 8.
\]
Now we can evaluate each part separately.
1. **Evaluating \(\sum_{n=1}^{80} 2n\)**:
\[
\sum_{n=1}^{80} 2n = 2 \sum_{n=1}^{80} n.
\]
The sum of the first \(k\) integers is given by the formula:
\[
\sum_{n=1}^{k} n = \frac{k(k+1)}{2}.
\]
Applying this for \(k = 80\):
\[
\sum_{n=1}^{80} n = \frac{80 \cdot 81}{2} = 40 \cdot 81 = 3240.
\]
Thus,
\[
\sum_{n=1}^{80} 2n = 2 \cdot 3240 = 6480.
\]
2. **Evaluating \(\sum_{n=1}^{80} 8\)**:
Since 8 is a constant, we can compute this sum as:
\[
\sum_{n=1}^{80} 8 = 8 \cdot 80 = 640.
\]
Now we can combine both results:
\[
\sum_{n=1}^{80} (2n - 8) = 6480 - 640 = 5840.
\]
Thus, the final result is
\[
\boxed{5840}.
\]
\[
\sum_{n=1}^{80} (2n - 8),
\]
we can simplify the sum by distributing it:
\[
\sum_{n=1}^{80} (2n - 8) = \sum_{n=1}^{80} 2n - \sum_{n=1}^{80} 8.
\]
Now we can evaluate each part separately.
1. **Evaluating \(\sum_{n=1}^{80} 2n\)**:
\[
\sum_{n=1}^{80} 2n = 2 \sum_{n=1}^{80} n.
\]
The sum of the first \(k\) integers is given by the formula:
\[
\sum_{n=1}^{k} n = \frac{k(k+1)}{2}.
\]
Applying this for \(k = 80\):
\[
\sum_{n=1}^{80} n = \frac{80 \cdot 81}{2} = 40 \cdot 81 = 3240.
\]
Thus,
\[
\sum_{n=1}^{80} 2n = 2 \cdot 3240 = 6480.
\]
2. **Evaluating \(\sum_{n=1}^{80} 8\)**:
Since 8 is a constant, we can compute this sum as:
\[
\sum_{n=1}^{80} 8 = 8 \cdot 80 = 640.
\]
Now we can combine both results:
\[
\sum_{n=1}^{80} (2n - 8) = 6480 - 640 = 5840.
\]
Thus, the final result is
\[
\boxed{5840}.
\]