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Use mathematical induction to show that when $n$ 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=$2^k$,for k >1} \end{cases}$

is $T(n) = n\ lg\ n$.

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