Evaluate the summation sigma-summation Underscript i equals 1 Overscript 12 EndScripts left-parenthesis i plus 3 right-parenthesis.

Question

(1 point)
Responses

15
15

78
78

81
81

114
114

Answer 1

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 \(\boxed{114}\).