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I've been having trouble with an assignment I received with the course I am following. The assignment in question:

Use induction to prove that when $n \geq 2$ is an exact power of $2$, the solution of the recurrence

$$T(n) = \begin{cases} 2 & \text{ if } n = 2,\\ 2T(n/2)+n & \text{ if } n =2^k, k > 1 \\ \end{cases} $$

is $T(n) = n\log(n)$

NOTE: the logarithms in the assignment have base $2$.

The base case here is obvious, when $n = 2$, we have that $2 = 2\log(2)$. However, I am stuck on the step here and I am not sure how to solve this.

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up vote 1 down vote accepted

Let $T(n)=n\log{n}$, here $n=2^k$ for some k. Then I guess we have to show that equality holds for $k+1$, that is $2n=2^{k+1}$. $T(2n)=2T(n)+2n=2n\log{n}+2n=2n(\log{n}+1)=2n\log{2n}$

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