Tower Of Hanoi Time Calculation
Starting from this system:

$$ \begin{cases} T(1) = 1 \\
 T(n) = 2T(n-1) + n \end{cases} $$

To use repertoire method to solve the system, we'll generalize it as this:

$$ \begin{cases} T(1) = \alpha \\
 T(n) = 2T(n-1) + \beta n + \gamma \end{cases} \tag 0 $$

The result must look like this:

$$ T(n) = A(n) \alpha + B(n) \beta + C(n) \gamma $$

I. Let's try T(n) = 1
$$ \begin{cases} T(1) = 1 = \alpha \\
 T(n) = 1 = 2T(n-1) + \beta n + \gamma = 2 \cdot 1 + \beta n + \gamma \end{cases}$$
Thus:
$$ (\alpha, \beta, \gamma) = (1, 0, -1) $$
Which means that 
$$ T(n) = A(n) - C(n) = 1  \tag 1 $$.

II. Now try T(n) = n
$$ \begin{cases} T(1) = 1 = \alpha \\
 T(n) = n = 2T(n-1) + \beta n + \gamma = 2n - 2 + \beta n + \gamma \end{cases}$$
Thus:
$$ (\alpha, \beta, \gamma) = (1, -1, 2) $$
Which means that 
$$ T(n) = A(n) - B(n) + 2 C(n) = n  \tag 2 $$.

III. Also, for $ (\alpha, \beta, \gamma) = (1, 0, 0) $:
$$\begin{cases} T(1) = 1 \\
T(n) = 2T(n-1) = 2^n \end{cases}$$
Which means that 
$$ T(n) = A(n) = 2^{n-1}  \tag 3$$.

From (1) and (3):
$$ 2^{n-1} - 1 = C(n) $$

Now from this equation and (2):
$$2^{n-1} - B(n) + 2 (2^{n-1} -1) = n$$
$$2^{n-1} - n + 2^n -2 = B(n) $$

Final result for $ (\alpha, \beta, \gamma) = (1, 1, 0) $:
$$ T(n) = A(n) + B(n) $$
$$ T(n) = 2^{n-1} + 2^{n-1} - n + 2^n -2 $$
$$ T(n) = 2^{n+1} - n - 2$$