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The recurrence relation given is: \[ T(n) = T\left(\frac{3n}{4}\right) + 1 \] To solve this numerically, let's expand the recurrence: 1. Expand the first few terms: \[ T(n) = T\left(\frac{3n}{4}\right) + 1 \] \[ T\left(\frac{3n}{4}\right) = T\left(\frac{9n}{16}\right) + 1 \] \[ T\left(\frac{9n}{16}\right) = T\left(\frac{27n}{64}\right) + 1 \] 2. Generalizing this pattern: \[ T(n) = T\left(\frac{3^k n}{4^k}\right) + _______ __________ ______ ____ _______ ____ __________ _____.
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