Determing growth rate of a sequence (1 Viewer)

[tacogym27101990]

Thanks, its really confusing me, i was thinking of making n =10^k to get rid of the down rounding operator but didnt really know where to go

 

[Affinity]

Thanks, its really confusing me, i was thinking of making n =10^k to get rid of the down rounding operator but didnt really know where to go

write k for floor(n^2/10)

then we know that a_n = k(k+1)/2

we also know that n^2/10 - 1 < k <= n^2/10


so
(n^2/10)(n^2/10 - 1)/2 < a_n <= (n^2/10)(n^2/10 + 1)/2

both bounds ~ n^4/200 and so does a_n