Q:

Recurrence Relation Homework Struggles

Here's the question:
Solve the recurrence by obtaining a theta bound for T(n) given that T(1) = theta(1).
T(n) = n + T(n-3)

Attempted Solution:
T(n) = T(n-6) + (n-3) + n  

= T(n-9) + (n-6) + (n-3) + n  

= T(n-(n-1)) + [(n-n) + (n-(n-3)) + (n-(n-6)) + ... + n]

= T(1) + [0 + 3 + 6 + ... + n]

= theta(1) = 3[1 + 2 + 3 + ... + n/3]

= theta(1) + [(n/3)(n/3 + 1)]/2

= theta(1) + (n^2+3n)/6

When I double check to see if the solution fits the recurrence, it doesn't work.

A:

The issue was that you were getting the wrong summation.
It doesn't start at 0, since your last T function was T(n - (n-1)) , which means previous one was T(n-(n-4)). So the summation starts at 4, and goes up till n.
If you don't know how to find the summation of this, I'd suggest you look at some of the proofs from the summation formula. This is what the solution looks like.
T(n) = T(n-3) + n  

= T(n-6) + (n-3) + n  

= T(n-(n-1)) + [ (n-(n-4)) + (n-(n-7)) + ... + n]

= T(1) + [4 + 7 + ... + n]

= theta(1) + (4 + n) * (n - 1)/6

