$$\mathrm{T}(n)=\begin{cases} 1 & n=1 \\ 3\cdot \mathrm{T}(n/4)+1 & n>1 \end{cases}$$ I got to: $$3^k\cdot \mathrm{T}(n/4)+3^{k-1}+3^{k-2}+\cdots+3+1$$
Can someone please help me?
2026-04-06 01:07:53.1775437673
How will you solve the following recurrence equation?
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