Menu Close

Prove-that-for-n-4-S-n-k-1-n-k-k-is-never-a-perfect-cube-




Question Number 187296 by anurup last updated on 15/Feb/23
Prove that for n≥4, S_n = Σ_(k=1) ^n k^(k!)  is never a perfect cube.
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{4},\:\mathrm{S}_{{n}} =\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{k}!} \:\mathrm{is}\:\mathrm{never}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{cube}. \\ $$
Commented by anurup last updated on 15/Feb/23
Please solve this.
$$\mathrm{Please}\:\mathrm{solve}\:\mathrm{this}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *