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?

By conditions of the problem:

(a+rs)(b + rs) = 100*T(n+s), for positive integers a, b, r and t with s=0, 1, 2, ..... where :

T(x) = xth triangular Number = x(x+1)/2

Thus,

ab = 100*T(n) = 50n(n+1)......(i)

(a+r)(b+t) = 100*T(n+1).......(ii)

(a+2r)(b+2t) = 100*T(n+2).....(iii)

(a+3r)(b+3t) = 100*T(n+2).....(iv); and so on

Thus, subtracting (i) from (ii); (ii) from (iii) and (iii) from (iv);

we have:

at + br + rt = 100(n+1)

at + br + 3rt = 100(n+2)

at + br + 5rt = 100(n+3)

Thus, 2rt = 100, giving rt = 50, so that:

at + 50b/t = 50(2n+1)........(v)

Or, (at - 50b/t)^2

= 2500(2n+1)^2 - 200ab

= 2500(2n+1)^2 - 10000(n^2+n)

= 2500

Or, at - 50b/t = +/-50.......(vi)

Solving for (v) and (vi), we have:

(at, 50b/t) = (50n, 50(n+1); (50(n+1), 50n)

Or, (50a/r, 50b/t) = (50n, 50(n+1); (50(n+1), 50n)

Or, (a/r, b/t) = (n, n+1); (n+1, n)

Or, (a, b) = (nr, (n+1)t), ((n+1)r, nt), with rt = 50

.........(*)

Consequently, (*) gives the generalised parametric solution to the given problem.