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+1 vote

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*no need to solve the recurrence relation just see the leading term in s1 and s2 .*

*In s1 as the value of r is increased the terms keep on decreasing due to exponential function in denominator. So the term where r =1 I.e n/2 will be the leading and thus it will be the order of n. *

*In S2 the last term where r=logn-1 is the leading term as exponential function is present in numerator so if you simplify it you would get Nolan.*

0 votes

*no need to solve it just see the leading term in s1 and s2 .*

*In s1 as the value of r is increased the terms keep on decreasing due to exponential function in denominator. So the term where r =1 I.e n/2 will be the leading and thus it will be the order of n. *

*In S2 the last term where r=logn-1 is the leading term as exponential function is present in numerator so if you simplify it you would get Nlogn.*

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