Solve the following recurrence: $E[T(n)] \leq \frac{2}{n} ∑_{k=\left\lfloor\frac{n}{2}\right\rfloor}^{n-1} E[T(k)] + 1$

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I am trying to solve the following recurrence but I do not really get how I could do it.

Basically the task is very similar to CLRS page. 216-219. I already found the expected number of recursive calls. It is:

$\frac{2}{n} ∑_{k=\left\lfloor\frac{n}{2}\right\rfloor}^{n-1} E[T(k)] + 1$

I got it the following way:

I now the recurrence for the time is:

$ T(n) \leq $ $∑_{k=1}^{n} X_k T(max[k,n-k])+ Θ(n)$

and since I am interested in the expected number of recursive calls we have that $Θ(n) = 1 :

$ T(n) \leq $ $∑_{k=1}^{n} X_k T(max[k,n-k])+ 1$

So far I am sure.

I take the expectation and to more or less the same calculations like in the book and get:

$E[T(n)] \leq \frac{2}{n} ∑_{k=\left\lfloor\frac{n}{2}\right\rfloor}^{n-1} E[T(k)] + 1$

Unfortunately here I am stuck. Does anyone now how to solve the recurrence?


When it come to the expected number of time, the recurrence is:

$T(n) \leq \frac{2}{n} ∑_{k=\left\lfloor\frac{n}{2}\right\rfloor}^{n-1} E[T(k)] + Θ(n)$

They show that $E(T(n)) = O(n))$ and start by assuming $E(T(n)) \leq cn$ for a constant $c$ by substition. I understand how they do it, but I do not see what I want to show by assuming.

Would be great if anyone could help me.