Prove that 1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1, for n >= 1
Here's what I have:
Base case: n = 1.
s(n) = 1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1
Therefore, (1+1)! - 1 = 1
Correct...
Inductive step:
For n+1,
or...
This is where I am stuck. What next?
Thank you so much for your help. I cannot thank you enough.
However, I am still confused as to how you reached the solution.
From what I see here, you factored out (n+1)!, correct? I follow at this point...
Okay, you lost me here. Can you please clarify this once more?