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+1

n

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....**

+1

@ Gupta731 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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