Solving Recurrence Relation T(n) = T(n-1) + n + 2 I am getting stuck solving this recurrence relation.
T(1) = 1
T(n) = T(n-1) + n + 2
I continued to expand it:
$T(n) = T(n-1) + n + 2 = [T(n-2) + (n-1) + n + 2] + 2$
$= T(n-2) + (n-1) + n +4 = [T(n-3) + (n-2) + (n-1) + n + 2] + 4$
$= T(n-3) + (n-2) + (n-1) + n + 6$
$= T(n-(n-1)) + (\sum_{i=0}^{n-2} n-i) + 2(n-1)$
$= 1 + (\sum_{i=0}^{n-2} n-i) + 2n-2$
This is where I get stuck. Did I do a step wrong?

HINT: 
$$T(n) - T(n-1) = n +2 = T(n-1) - T(n-2) + 1 \implies T(n) - 2T(n-1) + T(n-2) = 1$$
Now in the same manner:
$$T(n) - 2T(n-1) + T(n-2) = 1 = T(n-1) - 2T(n-2) + T(n-3) \implies $$
$$T(n) - 3T(n-2) + 3T(n-2) - T(n-3) = 0$$
Now this is a homogenous recursive relation, which can be easily solved.
