The following sequence gives a series of square numbers,
10% of 10=1,
20% 0f 20=4,
30% of 30=9,
40% of 40=16,........
90% 0f 90=81,
100% 0f 100=100,.....
Is it possible to create a similar series that gives us triangular numbers as the result?
(In reply to
re: of course  Similar Result by nikki)
I tried to prove this inductively, but I think it’s been a while since I did that kind of proof so I confused myself. However, I wonder if this will suffice:
I wanted to say that if I call the triangle numbers t1, t2, … tn then I have
10*n% of xn = tn
or
10*n/100 * xn = tn
tn = t[n1] + n = n*(n+1)/2
xn = x[n1] + 5 = 5*(n+1) [since I said x1=10]
Well,
10*n/100 * xn = tn = n*(n+1)/2
10/100*xn = (n+1)/2
xn = 100/10*(n+1)/2
xn = 5*(n+1)
Does this count as a proof?
Similarly, Ady’s pattern follows:
5*n% of yn = tn
or
5*n/100 * yn = tn
tn = t[n1] + n = n*(n+1)/2
yn = y[n1] + 10 = 10*(n+1) [since Ady said y1=20]
Well,
5*n/100 * yn = tn = n*(n+1)/2
5/100*yn = (n+1)/2
yn = 100/5*(n+1)/2
yn = 10*(n+1)

Posted by nikki
on 20050118 15:12:44 