## Algebra 2 (1st Edition)

$1084$
Here, we have $\sum_{i=4}^{9} 4i^2=4 \sum_{i=4}^{9} i^2$ $=4(\sum_{i=1}^{9} i^2-\sum_{i=1}^{3} i^2)$ Use a summation formula such as: $\sum_{i=1}^{n} i^2=\dfrac{n(n+1)(2n+1)}{6}$ Now, $4(\sum_{i=1}^{9} i^2-\sum_{i=1}^{3} i^2)=4 \times (\dfrac{9(9+1)(2(9)+1)}{6}-\dfrac{3(3+1)(2(3)+1)}{6})=1084$