Asked by Jaz
Evaluate the summation sigma-summation Underscript i equals 1 Overscript 12 EndScripts left-parenthesis i plus 3 right-parenthesis.
(1 point)
Responses
15
15
78
78
81
81
114
114
(1 point)
Responses
15
15
78
78
81
81
114
114
Answers
Answered by
GPT-4o mini
To evaluate the summation \(\sum_{i=1}^{12} (i + 3)\), we can break it down as follows:
1. Split the summation:
\[
\sum_{i=1}^{12} (i + 3) = \sum_{i=1}^{12} i + \sum_{i=1}^{12} 3
\]
2. Evaluate each part:
- The first part \(\sum_{i=1}^{12} i\) is the sum of the first 12 natural numbers. The formula for the sum of the first \(n\) natural numbers is:
\[
\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}
\]
So for \(n = 12\):
\[
\sum_{i=1}^{12} i = \frac{12(12 + 1)}{2} = \frac{12 \times 13}{2} = 78
\]
- The second part \(\sum_{i=1}^{12} 3\) is simply \(3\) added 12 times:
\[
\sum_{i=1}^{12} 3 = 3 \times 12 = 36
\]
3. Combine the two results:
\[
\sum_{i=1}^{12} (i + 3) = 78 + 36 = 114
\]
Thus, the final result of the summation is \(\boxed{114}\).
1. Split the summation:
\[
\sum_{i=1}^{12} (i + 3) = \sum_{i=1}^{12} i + \sum_{i=1}^{12} 3
\]
2. Evaluate each part:
- The first part \(\sum_{i=1}^{12} i\) is the sum of the first 12 natural numbers. The formula for the sum of the first \(n\) natural numbers is:
\[
\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}
\]
So for \(n = 12\):
\[
\sum_{i=1}^{12} i = \frac{12(12 + 1)}{2} = \frac{12 \times 13}{2} = 78
\]
- The second part \(\sum_{i=1}^{12} 3\) is simply \(3\) added 12 times:
\[
\sum_{i=1}^{12} 3 = 3 \times 12 = 36
\]
3. Combine the two results:
\[
\sum_{i=1}^{12} (i + 3) = 78 + 36 = 114
\]
Thus, the final result of the summation is \(\boxed{114}\).