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For $n \in \{1, 2, 3, 4\}$:  \[T(2^n) = n(5 \cdot 2^{2(n-1)})\]  Now we show that the above expression is true for all $n > \geq  0$: \begin{equation}  \begin{split}  T(2^{n+1}) &= 5 \cdot (2^n)^2 + (2^2)T(2^n)\\