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in Algorithms edited by
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T(n) = 10T(n/2) + 100n3

a=10, b= 2, k= 3 ,p=0

Here  a > b

So According to Masters theorem (Rule 1)-

T(n) = $\theta$( nlogba  )

            = $\theta$( nlog210)

         =  $\theta$(n3).           ( log210 >3)

Ans- A

         

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so 2^10=.2^(2^3.2)(that is greater than 3

            =n^3.2  > f(n)

so it will be ⦶(n^3)

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