You wrote:
i understand how to do ordinary induction proofs and i understand that strong induction proofs are the same as ordinary with the exception that you have to show that the theorem holds for all numbers up to and including some n (starting at the base case) then we try and show: theorem holds for $n+1$
No, not at all: in strong induction you assume as your induction hypothesis that the theorem holds for all numbers from the base case up through some $n$ and try to show that it holds for $n+1$; you don’t try to prove the induction hypothesis.
In your example the simple induction hypothesis that the result is true for $n$ is already enough to let you prove that it’s true for $n+1$, so there’s neither need nor reason to use a stronger induction hypothesis. The proof by ordinary induction can be seen as a proof by strong induction in which you simply didn’t use most of the induction hypothesis.
I suggest that you read this question and my answer to it and see whether that clears up some of your confusion; at worst it may help you to pinpoint exactly where you’re having trouble.
Added: Here’s an example of an argument that really does want strong induction. Consider the following solitaire ‘game’. You start with a stack of $n$ pennies. At each move you pick a stack that has at least two pennies in it and split it into two non-empty stacks; your score for that move is the product of the numbers of pennies in the two stacks. Thus, if you split a stack of $10$ pennies into a stack of $3$ and a stack of $7$, you get $3\cdot7=21$ points. The game is over when you have $n$ stacks of one penny each.
Claim: No matter how you play, your total score at the end of the game will be $\frac12n(n-1)$.
If $n=1$, you can’t make any move at all, so your final score is $0=\frac12\cdot1\cdot0$, so the theorem is certainly true for $n=1$. Now suppose that $n>1$ and the theorem is true for all positive integers $m<n$. (This is the strong induction hypothesis.) You make your first move; say that you divide the pile into a pile of $m$ pennies and another pile of $n-m$ pennies, scoring $m(n-m)$ points. You can now think of the rest of the game as splitting into a pair of subgames, one starting with $m$ pennies, the other with $n-m$.
Since $m<n$, by the induction hypothesis you’ll get $\frac12m(m-1)$ points from the first subgame. Similarly, $n-m<n$, so by the induction hypothesis you’ll get $\frac12(n-m)(n-m-1)$ points from the second subgame. (Note that the two subgames really do proceed independently: the piles that you create in one have no influence on what you can do in the other.)
Your total score is therefore going to be
$$m(n-m)+\frac12m(m-1)+\frac12(n-m)(n-m-1)\;,$$
which (after a bit of algebra) simplifies to $\frac12n(n-1)$, as desired, and the result follows by (strong) induction.
Best Answer
The reason you are confused is that (if I understand your problem correctly) $T(n)$ is defined only for $n = 2^k$, i.e. only when $n$ is a power of two.
Often when you do induction, you assume the statement is true for $n$, and then prove it is true for $n + 1$. But here, $T$ isn't even defined for $n + 1$ because it is only defined for powers of two. So you're going to have to be more clever.
You don't have to worry about $n + 1$ because it's not a power of two. What you do have to worry about is all the powers of two. So the way you would prove this by induction is, assuming it is true for $n$, prove it is true for the next power of two.
Letting $n = 2^k$, the next power of two is $2^{k+1}$. Therefore, you're going to want to assume that $$ T(n) = n \log_2 ( n ) \quad \textbf{for } \textbf{n = } \textbf{2}^\textbf{k} $$ and prove that $$ T(n) = n \log_2 (n) \quad \textbf{for } \textbf{n = } \textbf{2}^\textbf{k + 1}. $$