Let f(n)=an4+bn3+cn2+dn+e
f(0)=e=8;
f(1)=a+b+c+d+e=20, so a+b+c+d=12;
f(2)=16a+8b+4c+2d=64-8=56;
f(3)=81a+27b+9c+3d=125-8=117;
f(4)=256a+64b+16c+4d=162-8=154;
A=f(2)-2f(1)=14a+6b+2c=32, 7a+3b+c=16;
B=f(3)-3f(1)=78a+24b+6c=81, 26a+8b+2c=27;
C=f(4)-4f(1)=252a+60b+12c=106, 126a+30b+6c=53;
B-2A=12a+2b=-5;
C-3B=48a+6b=-28;
(C-3B)-3(B-2A)=12a=-13, a=-13/12; -13+2b=-5, b=4; -91/12+12+c=16, c=16+91/12-12=(192+91-144)/12=139/12; d=12-(-13/12+4+139/12)=12-29/2=-5/2.
f(n)=-13n4/12+4n3+139n2/12-5n/2+8.
So n=5 produces f(5)=108, the next term in the series according to the formula (all other terms fit).
But this is only one of several solutions.