Recurrence relation with sum
The trick here is to get rid of the sum. Note that the sum is up and down, it can be replaced by twice up:
$$
\begin{align*}
  (n + 1) T(n + 1) &amp;= 2 \sum_{1 \le k \le n} T(k) + c (n + 1)^2 \\
  n T(n)           &amp;= 2 \sum_{1 \le k \le n - 1} T(k) + c n^2 \\
  (n + 1) T(n + 1) - n T(n) &amp;= 2 T(n) + c (2 n + 1)
\end{align*}
$$
This is a linear, first order, non-homogeneous recurrence.

Update: Solution to the recurrence.

The recurrence can be written:
$$
\frac{T(n + 1)}{n + 2} - \frac{T(n)}{n + 1} = c \frac{2 n + 1}{(n + 1)(n + 2)}
$$
This reduces to a sum:
$$
\begin{align*}
\frac{T(n)}{n + 1} &amp;= T(0) + c \sum_{0 \le k \le n - 1}
                                 \frac{2 k + 1}{(k + 1) (k + 2)} \\
                   &amp;= T(0) + c \sum_{0 \le k \le n - 1}
                                 \left(
                                   \frac{3}{k + 2} - \frac{1}{k + 1}
                                 \right) \\
                   &amp;= T(0) + 3 c (H_{n + 1} - 1 - \frac{1}{2}) - c (H_n - 1) \\
                   &amp;= 2 c H_n + (T(0) - 2 c) + \frac{3c}{n + 1} \\
T(n)               &amp;= 2 c (n + 1) H_n + (T(0) - \frac{7 c}{2}) (n + 1) + 3 c \\
                   &amp;\sim 2 c n \ln n
\end{align*}
$$
Here $H_n$ is the $n$-th harmonic number:
$$
H_n = \sum_{1 \le k \le n} \frac{1}{k}
    = \ln n + \gamma + O(1/n)
$$



The recurrence $x_{n + 1} - a_n x_n = f_n$ can always be reduced to a telescoping sum at the left side by multiplying by the summing factor:
$$
(a_n a_{n - 1} \ldots a_0)^{-1}
$$
$$
\frac{x_{n + 1}}{a_n a_{n - 1} \ldots a_0} - \frac{x_n}{a_{n - 1} \ldots a_0}
   = \frac{f_n}{a_{n + 1} a_n \ldots a_0}
$$