Question Number 212991 by efronzo1 last updated on 28/Oct/24 | ||
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Answered by Frix last updated on 28/Oct/24 | ||
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$${H}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{k}!\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right) \\ $$$$\frac{{H}\left({n}\right)+\mathrm{1}}{\left({n}+\mathrm{1}\right)!}={n}+\mathrm{1}\:\:\:\:\:\left[\mathrm{test}\:\mathrm{it}\:\mathrm{for}\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},...\right] \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2026} \\ $$ | ||