## Algebra and Trigonometry 10th Edition

$6050$
$\sum_{k=1}^{10} 2k^3=2\sum_{k=1}^{10} k^3$ The sum of the cubes of the first $n$ integers can be obtained by the following formula: $\frac{n^2(n+1)^2}{4}.$ Thus the sum:$\frac{10^2(10+1)^2}{4}=\frac{10^211^2}{4}=\frac{12100}{4}=3025$ Thus $2\sum_{k=1}^{10} k^3=2\cdot3025=6050$