(In reply to
Arithmetic series by lucky)
Urgh, I complicated the proof unnecessarily......that's what happens when I am too tired to concetrate...
Anyway, the quoted part below was unneccessarily long:
"Also, if we observe the series carefully, we notice that the number of numbers in the series at any point equals:
n = (Rn/2) + 1
From this, we have that Rn = 2(n  1).
(To prove this point we can show the following series representing n:
1 + (2/2) + (2/2) + ... + (2/2) where its sum is
S = n = 1 + 2(n1)/2 (n1 is the number of (2/2)s in the series of n).
The series of Ri can be shown as:
R1, R1 + 2, R2 + 2, ... , Rn1 + 2 where R1 = 0.
From this we have that Rn = 2(n1). If we put this in the above sum, we have proved the point)"
All I needed to say instead of the above paragraphs is:
The series of Ri can be shown as:
R1, R2, R3, ... , Rn which corresponds to:
R1, R1 + 2, R2 + 2, ... , Rn1 + 2 (note that R1 = 0).
From this we have that Rn = R1 + 2(n1) = 2(n1).
sigh
btw, flooble seems to be overloaded lately...

Posted by lucky
on 20020808 10:32:46 