Solve the following recurrences (show all details): = 27 ( ² ) + m² (7) 9n b) T(n)=T 10 c) T(n)= T(n-1)+Ⓒ(n) a) T(n)=27 - (n)

pnp/1344587/mod ution.pdf 150% Q4) Solve the following recurrences using the Master Theorem. If it is not possible to solve it using the Master theorem then use Backward Substitution. a. T(n) = 4T () +1 b. T(n) = 3T (). + log2 n T(n) = T () + logn C.

Solve the following recurrences exactly:(a) T(1) = 8, and for all n ≥ 2, T(n) = 3T(n − 1) + 15.(b) T(1) = 1, and for all n ≥ 2, T(n) = 2T(n/2) + 6n − 1 (n is a power of 2)

Solve the recurrence relations shown below. a. T(n)={4T(n−1)+1if n>11if n=1T(n)={4T(n−1)+1if n>11if n=1 -- b. T(n)={3T(n/3)+4nif n>11if n=1

If lim n->oo (f(n))/(g(n)) = 0 then f(n) = ω( g(n) ) true false

Given the following recurrence, find the growth rate of T(n):T(n) = 4T(n/2) + 6n3 with T(1) = Θ(1)

Use the substitution method to show that the recurrence defined by T(n) = 2T(n/3) + Θ(n) hassolution T(n) = Θ(n).

Consider the following recurrence: T(1)=1; T(n) = 2.T()+n, for n> 1, n a power of 3. =..... Find T(27) by substitution, starting with n = 1, n = 65 2 169 29 15 6

We are given a recurrence of T(n) = T(n-1) + (n). For this recurrence it must be proven that T(n) is Ⓒ (n^2).

1. How to solve the following recurrence using backward substitution. O T(n) = /2 * T(n/2) + c, for n> 1 T(1) = a 0 (n²) o(n*) O(n log n) O (2")