How to solve $T(n)=\frac2n\sum_{i=1}^{n-1}T(i)+\Theta(n)$

Question

I need to solve this Recursion Relation:$$T(n)=\frac2n\sum_{i=1}^{n-1}T(i)+\Theta(n)$$ Do you have any idea how?
(I try to find another way instead induction).

(We should get: $T(n)=\Theta(n\log n)$)

This might help - [From the book we should see the solution]:

Thank you!

Answer 1

Let $$A(x) = \sum_{n\geq 0} T(n)\,x^n.$$ Then: $$\frac{A(x)}{1-x}=\sum_{n\geq 0}\left(\sum_{m\leq n}T(n)\right) x^n,\qquad A(x)\cdot\frac{x}{1-x}=\sum_{n\geq 0}\left(\sum_{m<n}T(m)\right) x^n,$$ so the recursion gives: $$ A(x)=2\int_{0}^{x}\frac{A(t)}{1-t}\,dt+\Theta\left(\frac{1}{1-x}\right),$$ or: $$ (1-x)B'(x) = 2B(x)+\Theta\left(\frac{1}{1-x}\right), $$ from which: $$ B(x) = \frac{x+C}{(1-x)^2},\qquad A(x)=\frac{x+C}{(1-x)^2}$$ and $T(n)\leq (C+1) n$.