since the terms of the sequence are all odd numbers (2k+1), and there are k^2 terms <= the kth odd number, look for something like
2[√n]+1
That sequence is
3,3,3, 5,5,5,5,5,5,5, ...
Looks like we have to adjust it by 1 to get
2[√(n-1)]+1
So, b+c+d = 2
No, I don't have an actual proof, but maybe you can work from here. I'll think on it some.
the non-
decreasing sequence of odd integers {a1, a2, a3, .
. .} = {1,3,3,3,5,5,5,5,5,...} each positive odd
integer k appears k times. it is a fact that there
are integers b, c, and d such that, for all positive
integers n, añ = b[√(n+c)] +d. Where [x] denotes the largest integer not exceeding x. Find the sum of b+c+d.
Show step too hard
1 answer