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....
@ Gupta731 we can think like this -
$\sum n ≈n^2$,$\sum n^2 ≈n^3$ ,$\sum n^3 ≈n^4$....$\sum n^5 ≈n^6$