How do you solve the following reccurance relation?
R(n)=R(n-1)+n?
Thanks.
I don't know anything about methods for solving recurance relations. I'm working on a problem right now and if I can solve the relation R(n)=R(n-1)+n with R(0)=2, R(1)=4, I'll be able to finish the problem.
I have this: R(n-1)=R(n-2)+n, I subtracted this from R(n)=R(n-1)+n and solved for R(n), getting R(n)=2R(n-1)-R(n-2).
Hello, Chris11!
Your initial values don't work . . . I'll modify them.
We can crank out the first few values of the sequence: .Solve the following reccurance: .
Take the difference of consecutive terms of the sequence,
. . then take the difference of the differences, and so on.
. .
We see that the second differences are constant.
. . Hence, the generating function is of the second degree, a quadratic.
The general quadratic function is: .
Use the first three values of the sequence and set up a system of equations:
. . . . . Then solve this system.
. .
. .
Substitute into [4]: .
Substitute into [1]: .
Therefore: .