Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$89,964$
Recall the formula: $\displaystyle \sum_{k=1}^{n}k^{3}=\left[\dfrac{n(n+1)}{2}\right]^{2}$ We can see that for the given sequence, the index does not start at 1. So, we will rewrite the given sequence as: $\sum_{k=4}^{24} k^{3}=$ (Terms from 4 to 24) = (Terms from 1 to 24) - (Terms from 1 to 3) We rewrite the sequence as stated above and apply the sum formulas: $\displaystyle \sum_{k=4}^{24} k^{3}= \sum_{k=1}^{24}k^{3}-\sum_{k=1}^{3}k^{3} \\= \displaystyle [\dfrac{24(24+1)}{2}]^{2}-[\dfrac{3(3+1)}{2}]^{2} \\=(300)^{2}-(6)^{2} \\=89,964$