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Question Number 154334 by alcohol last updated on 17/Sep/21

$$the{n}^{th}termof$$$$1,2,6,24,120........is?$$

Commented by talminator2856791 last updated on 17/Sep/21

$$\mathrm{joke}\mathrm{question}.$$

Answered by ARUNG_Brandon_MBU last updated on 17/Sep/21

$$u_n=n!$$

Answered by puissant last updated on 17/Sep/21

$$U_1=1!,U_2=2!,U_3=6!....\Rightarrow U_n=n!..$$

Answered by MJS_new last updated on 17/Sep/21

$$a_n=\frac{53}{24}{n}^4-\frac{81}{4}{n}^3+\frac{1627}{24}{n}^2-\frac{375}{4}n+45$$$$b_n=\frac{48825}{n^4}-\frac{410795}{4n^3}+\frac{1752787}{24n^2}-\frac{85501}{4n}+\frac{53213}{24}$$$$\mathrm{or}\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{one}\mathrm{with}n!$$$$c_n=n!+n^5-15n^4+85n^3-225n^2+274n-120$$

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