Given an integer
n (
n≠0), there are a finite number of sequences of consecutive integers whose terms add up to
n (If
n=25, then 3+4+5+6+7=25 is one such sequence with 5 terms).
a. Find an equation for the number of terms of the longest such sequence for any positive integer
n.
b. Find equations for the bounds (the first and last terms) of the longest such sequence for any positive integer
n.
Hint: Once you have an equation for the number of terms, and for the first term of the sequence, the last term is simply one less than their sum.
Hint 2: Ducks have absolutely nothing to do with the problem.
(In reply to
Picky of me.... by Brian Wainscott)
(a) requires n to be positive, but not all the terms of the sequence. So I'd say the longest sequence has 2n terms:
-(n-1) + -(n-2) + ... + 0 + 1 +...+(n-1)+n
(eg, for n=4 you have -3 + -2 + -1 + 0 + 1 + 2 + 3 + 4)
for (b), the first term is -(n-1) and the last term is n.
I'll have to think a bit about the proof that this is the longest such sequence, but it sounds right....
re: Picky of me....leads to solution?
