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2n^2 +nlogn =theta(n^2) ,I have done the lower bound..Like finding out the n0 and the constant.. Help me with the upper bound constant and n0
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take any value ≥2
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2n^2+ n logn <=3n^2

where n^2>=nlogn   or, n>=logn

So, n0 >=logn  ,c=3

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Btw I had to prove that it  can be theta bounded.. I cud solve the lower bound exactly but for upper bound I m not getting the equality and < than relation at the same time. :) Anyway thanks for ur effort..
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ok , now see :)
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