<s>[INST] Solving recurrence relation without initial condition Any idea on how I can approach this recurrence relation? It is very different to other questions I have encountered where there is only one term of $T(n)$ on the RHS, and the initial condition isn't given as well.
$$T(n)=\frac2n\big(T(0)+T(1)+\ldots+T(n-1)\big)+5n$$
for $n\ge 1$.
 [/INST] Use generating functions. Define $t(z) = \sum_{n \ge 0} T(n) z^n$, shift the recurrence and multiply out to get rid of fractions; multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
  (n + 1) T(n + 1)
     &= \sum_{0 \le k \le n} T(k) + 5 (n + 1)^2 \\
  \sum_{n \ge 0} (n + 1) T(n + 1) z^n
     &= \sum_{n \ge 0} z^n \sum_{0 \le k \le n} T(k)
          + 5 \sum_{n \ge 0} (n + 1)^2 z^n \\
   \frac{1}{z} \sum_{n \ge 0} n T(n) z^n
     &= \sum_{n \ge 0} z^n \sum_{0 \le k \le n} T(k)
          + 5 \frac{1 + z}{(1 - z)^3} \\
   \frac{1}{z} \cdot z t'(z)
     &= \frac{t(z)}{1 - z} + 5 \frac{1 + z}{(1 - z)^3}
\end{align*}$
No need for any initial values here. Solve for $t(z)$:
$\begin{align*}
  t(z)
    &= \frac{10}{(1 - z)^2} + 5 \frac{1}{1 - z} \ln \frac{1}{1 - z} + \frac{c}{1 - z}
\end{align*}$
Using the generalized binomial theorem and the generating function of the harmonic numbers:
$\begin{align*}
   (1 - z)^{-m}
     &= \sum_{n \ge 0} (-1)^n \binom{-m}{n} z^n \\
     &= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} z^n \\
  H_n
     &= \sum_{1 \le k \le n} \frac{1}{n} \\
   \sum_{n \ge 0} H_n z^n
     &= \frac{1}{1 - z} \ln \frac{1}{1 - z}
\end{align*}$
we see that:
$\begin{align*}
  T(n)
    &= [z^n] t(n) \\
    &= 10 \binom{n + 2 - 1}{2 - 1} + 5 H_n + c \\
    &= 10 n + 5 H_n + c'
\end{align*}$
Here $c' = T(0)$.
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