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Category: Relation and Functions

1-4-16-2-64-6-4-n-n-

Question Number 160136 by alcohol last updated on 25/Nov/21 $$\mathrm{1}+\mathrm{4}+\frac{\mathrm{16}}{\mathrm{2}}+\frac{\mathrm{64}}{\mathrm{6}}+…+\frac{\mathrm{4}^{{n}} }{{n}!}=? \\ $$ Answered by puissant last updated on 25/Nov/21 $${S}_{{n}} =\mathrm{1}+\mathrm{4}+\frac{\mathrm{16}}{\mathrm{2}}+\frac{\mathrm{64}}{\mathrm{6}}+….+\frac{\mathrm{4}^{{n}} }{{n}!}\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{4}^{{k}}…