Number of calls with length 2 array in quick-sort
First, note that you can write your recurrence as
$$
T_N = \frac{2}{N}\sum_{i=0}^{N-1} T_i.
$$
Consider now the generating function
$$
T(x) = \sum_{N=0}^\infty T_N x^N.
$$
We calculate
$$
\frac{T(x)}{1-x} = \sum_{N=0}^\infty \sum_{i=0}^N T_i x^N =
\frac{1}{2} \sum_{N=2}^\infty (N+1)T_{N+1}x^N = \frac{1}{2} \sum_{N=3}^\infty NT_Nx^{N-1}.
$$
On the other hand,
$$
\frac{d}{dx} T(x) = \sum_{N=0}^\infty NT_Nx^{n-1}.
$$
We conclude that $T(x)$ satisfies the differential equation
$$
\frac{d}{dx} T(x) = \frac{2T(x)}{1-x} + 2x.
$$
Additionally, $T(0) = 0$. Plugging this into Wolfram alpha, we obtain the solution
$$
T(x) = \frac{x^2(3x^2-8x+6)}{6(1-x)^2}.
$$
Using $1/(1-x)^2 = \sum_{N=0}^\infty (N+1)x^N$, we obtain for $N \geq 3$ that
$$
T_N = \frac{6(N-1)-8(N-2)+3(N-3)}{6} = \frac{N+1}{6}.
$$



Having computed the solution by brute force, it is easy to prove by induction. Suppose that $N \geq 3$. Then
$$
T_N = \frac{2}{N} \sum_{i=0}^{N-1} T_i =
\frac{2}{N} \left(1 + \sum_{i=3}^{N-1} \frac{i+1}{6}\right) =
\frac{2}{N} \left(1 + \frac{(N-3)(N+4)}{12} \right) = \frac{N+1}{6}.
$$