Posted by Soroban on September 10, 2002 at 00:00:53:
In Reply to: Quadratics posted by kimmay on September 09, 2002 at 18:40:11:
: I have a sequence of terms and am required to come up with a function.. apparently quadratic.. need help..
: {1,3,6,10,15,21,28,36,45...}
: thanks
Here is a "primitive" method of determining the generating function.
If you suspect that it's a quadratic function (and you're correct),
then f(n) = An^2 + Bn + C, and we must determine coefficients A,B,C.
Use the first three values from the sequence: f(1)=1, f(2)=3, f(3)=6.
This gives us three equations:
A + B + C = 1, 4A + 2B + C = 3, 9A + 3B + C = 6
which has solutions: A = 1/2, B = 1/2, C = 0.
The quadratic is (1/2)n^2 + (1/2)n, so f(n) = n(n + 1)/2.