Assume the nth term of the series is given by the polynomial f(n)=a₄n⁴+a₃n³+a₂n²+a₁n+a₀ where n starts at 0.
Therefore a₀=2 because f(0)=2, the first term.
Now subtract the first term a₀ from each of the others to reduce the series:
3, 8, 15, 18 for f(1), f(2), f(3), f(4).
We have 4 equations and 4 unknowns a₁—a₄:
① f(1)=3=a₄+a₃+a₂+a₁
② f(2)=8=16a₄+8a₃+4a₂+2a₁
③ f(3)=15=81a₄+27a₃+9a₂+3a₁
④ f(4)=18=256a₄+64a₃+16a₂+4a₁
Now we reduce the number of unknowns to three:
⑤ f(2)-2f(1)=2=14a₄+6a₃+2a₂⇒1=7a₄+3a₃+a₂
⑥ f(3)-3f(1)=6=78a₄+24a₃+6a₂⇒1=13a₄+4a₃+a₂
⑦ f(4)-4f(1)=6=252a₄+60a₃+12a₂⇒1=42a₄+10a₃+2a₂
⑧ ⑥-⑤=0=6a₄+a₃, so a₃=-6a₄
⑨ ⑦-2⑥=-1=16a₄+2a₃, so, substitute for a₃: -1=16a₄-12a₄=4a₄, a₄=-¼, making a₃=3/2.
Substitute these values in ⑤: 1=-7/4+9/2+a₂, making a₂=-7/4.
Substituting all these values in ①: a₁=3-(-7/4+3/2-1/4)=7/2.
Therefore f(n)=-x⁴/4+3x³/2-7x²/4+7x/2+2.
From this formula we get the series: 2, 5, 10, 17, 20, 7, -40, -145, -338, -655, -1138, ...