Time complexity for T(n-1)

Question

I am confused on solving this time complexity problem.

T(n) = T(n-1)

I know in quick-sort worst case T(n) = T(n-1) + T(1) + n

Which evaluates to (n-1) + (n-2) + (n-3) + ... + 1 & this geometric sequence equals O(n^2)

However. I see answers on stackoverflow that say T(n) = T(n-1) + c = O(n).

How is this possible when this is also equal to (n-1) + (n-2) + (n-3) + ... + 1, which equals O(n^2)

Can somebody please explain.

Answer 1

T(n) = T(n-1) + c isn't equal to (n-1) + (n-2) + (n-3) + ... + 1 because the terms being added are constants. Basically:

Adding nothing:

T(n) = T(n-1)
0 + 0 + 0 + ... + 0 = 0
O(1)

Adding a constant:

T(n) = T(n-1) + c
c + c + c + ... + c = nc
O(n)

Adding a variable:

T(n) = T(n-1) + n
1 + 2 + 3 + ... + n = n(n+1)/2
O(n^2)