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?

Triangular numbers are defined recursively:
T(n) = T(n-1) + n

So we want to prove that (.1*n)*(5n+5) = T(n)
(ie, that 10% of 10, 20% of 15, 30% of 20, 40% of 25, etc are all consecutive triangular numbers)

This is easily done inductively. The base case is obvious: T(1) = .1*(5+5) = 1.

By definition, T(n+1) = T(n) + n + 1.
And if we assume that T(n) = (.1n)(5n+5) = .5n^2 + .5n, then T(n+1) = .5n^2 + 1.5n + 1.

And using the formula we're trying to prove:
T(n+1) = (.1(n+1))(5(n+1)+5) = (.1n + .1)(5n + 10) = .5n^2 + 1.5n + 1

So since the base case works and the nth case implies the (n+1)th case, the formula T(n) = .1n(5n+5) is a true formula for any triangular number.

Posted by athar on 2005-02-08 03:25:06