Recurance Relationship involving factorials. (Discrete Math.)?

Question

Say i have the following sequence:
Note: A(n) = A subscript n. So A1 -> A subscript 1

A1 = 1(1!)
A2 = 1(1!) + 2(2!)
A3 = 1(1!) + 2(2!) + 3(3!)
hence:
A4 = A3 + 4(4!)
hence:
A(n) = A(n-1) + n(n!)

I know how to solve recurrance relationships in first or secondorder, but i've never come across a relationship that involves factorials. Need help in finding a general solution to this recurrance relationship to find the answer to an arbitary N-th term.

i.e. i need an equaltion to find: A(n) without having to recurrsively calculate A(n-1), A(n-2), A(n-3) ... etc.

The prove my induction shows that it's the correct answer, but how did you come up with that answer in the first place?

Thx

Answer 1

A[n] = (n + 1)! - 1

Proof:
The relationship holds for n = 1:
(1 + 1)! - 1 = 1(1!) = 1

If the relationship holds for n, then it also holds for n+1:
Assume that A[n] = (n+1)! - 1
By definition of A[n], we have:
A[n+1] = A[n] + (n+1) (n+1)!
Using the assumption above, we get:
A[n+1] = (n+1)! - 1 + (n+1) (n+1)!
= (n+1)! + (n+1) (n+1)! - 1
= (n+1)! (1 + n + 1) - 1
= (n+1)! (n+2) - 1
= (n+2)! - 1
So...
A[n+1] = (n+2)! - 1
That completes the proof.

Answer 2

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