Final answer:
The first recursion equation can be solved using the Master Theorem, which gives a solution of Theta(n^3 * log n). The second recursion equation can be solved using substitution and the formula for the sum of the first n perfect cubes, resulting in a solution of Theta(n^5).
Step-by-step explanation:
The first recursion equation, (1), can be solved using the Master Theorem. The equation can be rewritten as T(n) = 27T(n/3) + n^4. According to the Master Theorem, if the equation is of the form T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is an asymptotically positive function, then the solution can be determined by comparing the function f(n) to n^(log base b of a).
In this case, we have a = 27, b = 3, and f(n) = n^4. The function f(n) is polynomial, and n^(log base b of a) is a polynomial as well. Therefore, the solution to the equation is T(n) = Theta(n^(log base b of a) * log n) = Theta(n^3 * log n).
For the second recursion equation, (2), we can use substitution to solve it. We rewrite the equation as T(n) = T(n-1) + n^4. Substituting n with n-1 yields T(n-1) = T(n-2) + (n-1)^4. By repeatedly substituting, we can express T(n) in terms of T(1) and the base case of n^4 as follows: T(n) = T(1) + 1^4 + 2^4 + ... + (n-1)^4 + n^4. Using the formula for the sum of the first n perfect cubes, we can simplify the expression to T(n) = T(1) + n(n+1)(2n+1)(6n^2 + 6n + 1)/24. Therefore, the solution to the equation is T(n) = Theta(n^5).