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Solve using Masters theorem 

2T (n/2) + n log n

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we know,

   T(n) = a t(n/b) + Θ(n.logpn)
a=2, b=2 ,k=1 and p=1
so the condition is a=bk

and p>=-1
so  T(n)=(nlog 22)*log p+1n
so it is   n*log2n

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here a=2,b=2,k=1,p=1;

now,

here a=b

so we will use the formula T(n)=θ(n^log a to the base b log n^p+1)

so answer is θ(n log^2n)

 

here we will use the formula for masters theorem as

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