## Algebra and Trigonometry 10th Edition

$86100$
The sum of the squares of the first $n$ integers can be obtained by the following formula: $\frac{n(n+1)(2n+1)}{6}.$ Thus $\sum_{i=1}^{50} (2i^2+5)=\sum_{i=1}^{50} (5)+\sum_{i=1}^{50} (2i^2)=5\cdot50+2\sum_{i=1}^{50} (i^2)=250+2\cdot\frac{50(50+1)(2(50)+1)}{6}=250+2\cdot\frac{50(50+1)(2(50)+1)}{6}=250+85850=86100$