Given the pattern: T1 = 0, T2 = 3, T3 = 8, T4 = 15. Use the formula Tn = ___________ to find T10.

The given terms are:
$T_1 = 0$
$T_2 = 3$
$T_3 = 8$
$T_4 = 15$

First differences:
$T_2 - T_1 = 3 - 0 = 3$
$T_3 - T_2 = 8 - 3 = 5$
$T_4 - T_3 = 15 - 8 = 7$

Second differences:
$5 - 3 = 2$
$7 - 5 = 2$

Since the second differences are constant, the sequence is quadratic, meaning the general term $T_n$ can be expressed in the form $T_n = An^2 + Bn + C$.

The second difference is equal to $2A$.

$$2A = 2 \implies A = 1$$

Now substitute $A=1$ into the general formula: $T_n = n^2 + Bn + C$.

Use the first two terms to form a system of equations:
For $n=1$:

$$T_1 = 1^2 + B(1) + C = 0$$ $$1 + B + C = 0 \quad (1)$$

For $n=2$:

$$T_2 = 2^2 + B(2) + C = 3$$ $$4 + 2B + C = 3 \quad (2)$$

From equation (1), $B + C = -1$.
From equation (2), $2B + C = -1$.

Subtract equation (1) from equation (2):

$$(2B + C) - (B + C) = -1 - (-1)$$ $$B = 0$$

Substitute $B=0$ into equation (1):

$$1 + 0 + C = 0$$ $$C = -1$$

Substitute $A=1$, $B=0$, and $C=-1$ into $T_n = An^2 + Bn + C$:

$$T_n = 1n^2 + 0n - 1$$ $$T_n = n^2 - 1$$

Substitute $n=10$ into the formula for $T_n$:

$$T_{10} = (10)^2 - 1$$ $$T_{10} = 100 - 1$$ $$T_{10} = 99$$

The formula for $T_n$ is $\boxed{n^2 - 1}$.
The value of $T_{10}$ is $\boxed{99}$.