To solve the given problem, we'll need to evaluate the sum of the series [tex]\( \sum_{n=1}^3 (4n^2 + 3) \)[/tex].
1. Find the individual terms of the series:
- When [tex]\( n = 1 \)[/tex]:
[tex]\[
4(1)^2 + 3 = 4 \cdot 1 + 3 = 4 + 3 = 7
\][/tex]
- When [tex]\( n = 2 \)[/tex]:
[tex]\[
4(2)^2 + 3 = 4 \cdot 4 + 3 = 16 + 3 = 19
\][/tex]
- When [tex]\( n = 3 \)[/tex]:
[tex]\[
4(3)^2 + 3 = 4 \cdot 9 + 3 = 36 + 3 = 39
\][/tex]
2. List the terms of the series:
[tex]\[
7, 19, 39
\][/tex]
3. Sum the terms:
[tex]\[
7 + 19 + 39 = 65
\][/tex]
Therefore, the sum of the series [tex]\( \sum_{n=1}^3 (4n^2 + 3) \)[/tex] is [tex]\( 65 \)[/tex].
Enter your answer in the box:
[tex]\[
\boxed{65}
\][/tex]
