One solution is T(n)=(n-10)(n²-2n+3)/6.
Here’s why.
If T(n)=a₀+a₁n+a₂n²+a₃n³ where the a’s are constant coefficients and the first term is when n=0. We have to find a₀ to a₃. We have 4 coefficients and 4 equations controlled by the given terms.
T(0)=a₀=-3.
T(1)=-3+a₁+a₂+a₃=-4, so a₁+a₂+a₃=-1 (1)
T(2)=-3+2a₁+4a₂+8a₃=-7, so 2a₁+4a₂+8a₃=-4 (2)
T(3)=-3+3a₁+9a₂+27a₃=-11, so 3a₁+9a₂+27a₃=-8 (3)
T(2)-2T(1)=2a₂+6a₃=-2 (4)
T(3)-3T(1)=6a₂+24a₃=-5 (5)
(5)-3(4)=6a₃=1, a₃=⅙; (4) 2a₂+1=-2, a₂=-3/2; (1) a₁-3/2+1/6=-1, a₁=⅓.
T(n)=-3+n/3-3n²/2+n³/6=
⅙(-18+2n-9n²+n³)=
⅙(n(n²+2)-9(n²+2))=
⅙(n-9)(n²+2) for n≥0.
For n starting at 1, n-1 starts at 0, so T(n)=⅙(n-1-9)((n-1)²+2).
Therefore T(n)=⅙(n-10)(n²-2n+3) for n≥1. Note that this may not be the only solution.