Q6. Find the order of growth for solution of the following recurrence $T(n) = 4T(n/2) + n^2$, $T(1) = 1$ using the backward substitution method. (15)
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Each time we substitute back, the coefficient of T(n/2^k) is increasing by a factor of 4, and the exponent of n is decreasing by a factor of 2. So, after k substitutions, we will have: T(n) = 4^k T(n/2^k) + (n/2^k)^2 + (n/2^(k-1))^2 + ... + (n/2)^2 + n^2 Now, Show more…
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