Don't copy the answer from somewhere else. thank you

(a) Given the equation, T(n)=2T(n/2)+n, guess a solution of the form: T(n)=c1n+c2nlog2(n). Find the coefficients c1,c2 to determine the exact solution assuming a value T(1) at the bottom the recursion. (b) Generalize this to the case to the equation T(n)=aT(n/b)+nk and guess the solution of the form: T(n)=c1n+c2nk using b=a and assuming =k. First show if you drop the nk using the homogeneous equation T(n)=aT(n/b) the form c1n is a solution! (What is c1 ?) Second drop aT(n/b) and show c2nk is a solution (What is c2 ?) With both terms (and =k ) the full solution is just the sum of the to terms but only one or the other dominates! (c) What happens when the two solution collide (i.e have the same power, i.e = log(a)/log(b)=k.) Now show that the leading solution is as n goes to infinity is T(n)=(nklogn)1 1 If you are ambitious you find exact solutions for part b and c above are of the form T(n)=c1n+givec2nk and T(n)=c1nk+c2nlog(n)nk explicitly determine the cs for each case respectively. We did that in part for example. BUT note we already know the leading terms for the Master Equation without this extra effort! Neat trick