Question Number 154334 by alcohol last updated on 17/Sep/21
$${the}\:{n}^{{th}} \:{term}\:{of} \\ $$$$\mathrm{1}\:,\:\mathrm{2},\:\mathrm{6},\:\mathrm{24},\:\mathrm{120}\:……..\:{is}? \\ $$
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
$${U}_{\mathrm{1}} =\mathrm{1}!\:,\:\:{U}_{\mathrm{2}} =\mathrm{2}!\:,\:{U}_{\mathrm{3}} =\mathrm{6}!\:….\:\Rightarrow\:{U}_{{n}} =\:{n}!.. \\ $$
Answered by MJS_new last updated on 17/Sep/21
$${a}_{{n}} =\frac{\mathrm{53}}{\mathrm{24}}{n}^{\mathrm{4}} −\frac{\mathrm{81}}{\mathrm{4}}{n}^{\mathrm{3}} +\frac{\mathrm{1627}}{\mathrm{24}}{n}^{\mathrm{2}} −\frac{\mathrm{375}}{\mathrm{4}}{n}+\mathrm{45} \\ $$$${b}_{{n}} =\frac{\mathrm{48825}}{{n}^{\mathrm{4}} }−\frac{\mathrm{410795}}{\mathrm{4}{n}^{\mathrm{3}} }+\frac{\mathrm{1752787}}{\mathrm{24}{n}^{\mathrm{2}} }−\frac{\mathrm{85501}}{\mathrm{4}{n}}+\frac{\mathrm{53213}}{\mathrm{24}} \\ $$$$\mathrm{or}\:\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{one}\:\mathrm{with}\:{n}! \\ $$$${c}_{{n}} ={n}!+{n}^{\mathrm{5}} −\mathrm{15}{n}^{\mathrm{4}} +\mathrm{85}{n}^{\mathrm{3}} −\mathrm{225}{n}^{\mathrm{2}} +\mathrm{274}{n}−\mathrm{120} \\ $$