Just adding a bit clarification on how to jump from a massive DP to a clean nice concise mathematical formula.

Instead of looking at the result, let us explain the C(n,0) differently.

from the start

DP[2][2] = C(n,0) + C(n,1)DP[1][1] + C(n,2)DP[2][1]
while
(n+1)^2 = C(n,0) * n^0 * 1^2 + + C(n,1) * n^1 * 1^1 + C(n,2) * n^2 * 1 ^ 0;

We now could see that the C(n,0) means pick 0 n of (n + 1) and 2 1, C(n,1) means pick 1 n and 1 1.

thus give us the formula DP[2][2] = DP[1][2] ^ 2 = (t+1) ^ 2 ..... (t+1)^n