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C?
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Yes C is correct, please explain.
1

  n
sum i^5 ≈ n^6
k=1

and log n^40 = 40logn 

and n^6 >> 40logn

so x ≈ √ O(n^6) = O(n^3) = θ(n^3)= Ω(n^3)

also X= O(n^3) or O(n^4) or O(n^c) ..... c>=3....

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I am getting it but why $\sum$$i^5$ $\approx$ $n^6$
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since, ∑i^5= (1/6)n6 + (1/2)n5 + (5/12)n4 - (1/12)n2...
1

 we can think like this -

$\sum n ≈n^2$,$\sum n^2 ≈n^3$ ,$\sum n^3 ≈n^4$....$\sum n^5 ≈n^6$

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yeah got it..thanks

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