## Algebra and Trigonometry 10th Edition

$979$
The sum of the fourth powers of the first $n$ integers can be obtained by the following formula: $\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.$ Thus the sum:$\frac{5(5+1)(2( 5)+1)(3(5)^2+3(5)-1)}{30}=\frac{5(6)(11)(89)}{30}=11\cdot89=979$