Asked by fui
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
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
Answers
Answered by
Steve
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!