2-12-18-48-50-

Facebook Tweet Pin

Question Number 218970 by Nicholas666 last updated on 17/Apr/25

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2},\mathrm{12},\mathrm{18},\mathrm{48},\mathrm{50},….. \\ $$$$ \\ $$

Answered by Frix last updated on 17/Apr/25

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}\:\mathrm{as}\:\mathrm{always}. \\ $$

Commented by Nicholas666 last updated on 17/Apr/25

$${how}\:{do}\:{you}\:{solve}? \\ $$

Commented by Rasheed.Sindhi last updated on 19/Apr/25

a(n)= { ((2n^2 for an odd n)),((((−21n^2 +198n−264)/4) for an even n)) :} Next term is a(6)=((−21(6^2 )+198(6)−264)/4)=42

$$ \\ $$$${a}\left({n}\right)=\begin{cases}{\mathrm{2}{n}^{\mathrm{2}} \:{for}\:{an}\:{odd}\:{n}}\\{\frac{−\mathrm{21}{n}^{\mathrm{2}} +\mathrm{198}{n}−\mathrm{264}}{\mathrm{4}}\:{for}\:{an}\:{even}\:{n}}\end{cases} \\ $$$${Next}\:{term}\:{is} \\ $$$${a}\left(\mathrm{6}\right)=\frac{−\mathrm{21}\left(\mathrm{6}^{\mathrm{2}} \right)+\mathrm{198}\left(\mathrm{6}\right)−\mathrm{264}}{\mathrm{4}}=\mathrm{42} \\ $$

Commented by Frix last updated on 19/Apr/25

See? I told you so ��

Commented by Rasheed.Sindhi last updated on 19/Apr/25

100% true sir! ☺️

Answered by Rasheed.Sindhi last updated on 19/Apr/25

2,12,18,48,50,..... { ((2n^2 ;n∈{1,3,5,...})),((3n^2 ;n∈{2,4,6,...})) :} OR 2,12,18,48,50,..... 2n^2 (((1+(−1)^(n+1) )/2))+3n^2 (((1+(−1)^n )/2)) OR 2,12,18,48,50,....(n^2 /2)(5+(−1)^n ) Next two terms are: 108 and 98

$$\:\:\:\:\:\:\:\:\mathrm{2},\mathrm{12},\mathrm{18},\mathrm{48},\mathrm{50},…..\begin{cases}{\mathrm{2}{n}^{\mathrm{2}} \:\:\:\:;{n}\in\left\{\mathrm{1},\mathrm{3},\mathrm{5},…\right\}}\\{\mathrm{3}{n}^{\mathrm{2}} \:\:\:\:;{n}\in\left\{\mathrm{2},\mathrm{4},\mathrm{6},…\right\}}\end{cases}\: \\ $$$$\mathrm{OR} \\ $$$$\:\:\:\:\:\:\mathrm{2},\mathrm{12},\mathrm{18},\mathrm{48},\mathrm{50},…..\:\mathrm{2}{n}^{\mathrm{2}} \left(\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}}\right)+\mathrm{3}{n}^{\mathrm{2}} \left(\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\right) \\ $$$$\mathrm{OR} \\ $$$$\mathrm{2},\mathrm{12},\mathrm{18},\mathrm{48},\mathrm{50},….\frac{{n}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{5}+\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$${Next}\:{two}\:{terms}\:{are}:\:\mathrm{108}\:{and}\:\mathrm{98} \\ $$

Commented by Nicholas666 last updated on 19/Apr/25

$${thanks} \\ $$

Answered by Ghisom last updated on 19/Apr/25

zillions of possibilities... polynomial solution: a_n =−((10)/3)x^4 +38x^3 −((440)/3)x^2 +254x−120 or just continue: 2, 12, 18, 48, 48+2=50 50+12=62 62+18=80 80+48=128 128+2=130 ...

$$\mathrm{zillions}\:\mathrm{of}\:\mathrm{possibilities}… \\ $$$$\mathrm{polynomial}\:\mathrm{solution}: \\ $$$${a}_{{n}} =−\frac{\mathrm{10}}{\mathrm{3}}{x}^{\mathrm{4}} +\mathrm{38}{x}^{\mathrm{3}} −\frac{\mathrm{440}}{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{254}{x}−\mathrm{120} \\ $$$$\mathrm{or}\:\mathrm{just}\:\mathrm{continue}: \\ $$$$\mathrm{2},\:\mathrm{12},\:\mathrm{18},\:\mathrm{48}, \\ $$$$\mathrm{48}+\mathrm{2}=\mathrm{50} \\ $$$$\mathrm{50}+\mathrm{12}=\mathrm{62} \\ $$$$\mathrm{62}+\mathrm{18}=\mathrm{80} \\ $$$$\mathrm{80}+\mathrm{48}=\mathrm{128} \\ $$$$\mathrm{128}+\mathrm{2}=\mathrm{130} \\ $$$$… \\ $$

Answered by Rasheed.Sindhi last updated on 19/Apr/25

a(n)= { ((2n^2 for an odd n)),((2 ((n/2)+2)! for an even n)) :} Next two numbers are: 240 and 98

$${a}\left({n}\right)=\begin{cases}{\mathrm{2}{n}^{\mathrm{2}} \:\:\:{for}\:{an}\:{odd}\:{n}}\\{\mathrm{2}\:\left(\frac{{n}}{\mathrm{2}}+\mathrm{2}\right)!\:{for}\:{an}\:{even}\:{n}}\end{cases} \\ $$$${Next}\:{two}\:{numbers}\:{are}:\:\mathrm{240}\:{and}\:\mathrm{98} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply