Here is an explaination for the solution http://techieme.in/solving-recurrences-master-method/
T(n) = 7T(n/2) + n^2
comparing with the equation (MASTER THEOREM)
we get ,a=7,b=2,k=2,p=0
now it satisfies a>b^k,
so case first of master theorem
T(n)=nlog27 =n2.81 i.e f(n) is polynomially smaller than t(n), therefore it is first case
therefore, ans is ⊖(nlog7) or ⊖(n2.81)
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