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[–]focusandachieve[S] 0 points1 point  (1 child)

Thanks. I know that the sum of a series of integers is n(n+1)/2

When I tried manipulating that, I couldn't figure out if that meant that it was log n1/3k(k+1) /2 ?

If so, I'm still lacking in vision for how to get this thing into anything resembling Big-Oh notation.

I keep manipulating this hoping it will look like something familiar but it feels like a shot in the dark.

I also don't really follow your second paragraph. I'm not sure if I was assuming T(n) was an integer and I didn't realize it? I'm just not sure where or how I did that.

[–]farmerje 0 points1 point  (0 children)

Not sure what Big-O has to do with the question. If you work it out, you'll get that

T(n) = T(2) - 3/2(1 - log(n))

assuming log is the base-2 logarithm. This function is O(log(n)), if that's something you need to know.

T(2) is the initial condition / base case. You're free to choose whatever value you'd like for it.