Evaluate sigma-summation Underscript lower i equals 1 overscript 12 EndScripts left parenthesis 2 lower I squared minus 7 right parenthesis.

Question

(1 point)
Responses

281
281

566
566

1,216
1,216

1,293

Answer 1

To evaluate the summation \(\sum_{i=1}^{12} \left(2i^2 - 7\right)\), we can break it down into two parts:

1. Calculate \(\sum_{i=1}^{12} 2i^2\)
2. Calculate \(\sum_{i=1}^{12} 7\)

Step 1: Calculate \(\sum_{i=1}^{12} 2i^2\)

First, we need to calculate \(\sum_{i=1}^{12} i^2\). There is a formula for the sum of squares:

\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]

For \(n = 12\):

\[ \sum_{i=1}^{12} i^2 = \frac{12(12+1)(2 \times 12 + 1)}{6} = \frac{12 \times 13 \times 25}{6} \]

Calculating this step-by-step:

\[ 12 \times 13 = 156 \]

\[ 156 \times 25 = 3900 \]

\[ \frac{3900}{6} = 650 \]

Thus,

\[ \sum_{i=1}^{12} i^2 = 650 \]

Now multiply this result by 2:

\[ \sum_{i=1}^{12} 2i^2 = 2 \times 650 = 1300 \]

Step 2: Calculate \(\sum_{i=1}^{12} 7\)

This is simply:

\[ \sum_{i=1}^{12} 7 = 7 \times 12 = 84 \]

Step 3: Combine the Results

Now we subtract the second summation from the first:

\[ \sum_{i=1}^{12} (2i^2 - 7) = 1300 - 84 = 1216 \]

Thus, the value of the summation is:

\[ \boxed{1216} \]