# Thread: Need help proving mathematical induction

1. ## Need help proving mathematical induction

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,
$n(n!) = (n+1)*(n+1)!$

$s(n+1) = s(n) + (n+1)*(n+1)!$
or...
$s(n+1) = (n+1)! - 1 + (n+1)*(n+1)!$

This is where I am stuck. What next?

2. Originally Posted by kalel918
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,
$n(n!) = (n+1)*(n+1)!$

$s(n+1) = s(n) + (n+1)*(n+1)!$
or...
$s(n+1) = (n+1)! - 1 + (n+1)*(n+1)!$

This is where I am stuck. What next?
$(n+1)! - 1 + (n+1)*(n+1)! = (n+1)! (1 + [n+1]) - 1 = (n+1)!(n+2) - 1 = (n+2)! - 1$

3. Originally Posted by mr fantastic
$(n+1)! - 1 + (n+1)*(n+1)! = (n+1)! (1 + [n+1]) - 1 = (n+1)!(n+2) - 1 = (n+2)! - 1$
Thank you so much for your help. I cannot thank you enough.

However, I am still confused as to how you reached the solution.

$(n+1)! (1 + [n+1]) - 1$

From what I see here, you factored out (n+1)!, correct? I follow at this point...

$= (n+1)!(n+2) - 1 = (n+2)! - 1$

Okay, you lost me here. Can you please clarify this once more?

4. Originally Posted by kalel918
Thank you so much for your help. I cannot thank you enough.

However, I am still confused as to how you reached the solution.

$(n+1)! (1 + [n+1]) - 1$

From what I see here, you factored out (n+1)!, correct? I follow at this point...

$= (n+1)!(n+2) - 1 = (n+2)! - 1$

Okay, you lost me here. Can you please clarify this once more?
$(n + 2)! = (n+2) {\color{red}(n+1)n(n-1) ..... 1} = (n+2) {\color{red}(n+1)!}$.

5. Originally Posted by mr fantastic
$(n + 2)! = (n+2) {\color{red}(n+1)n(n-1) ..... 1} = (n+2) {\color{red}(n+1)!}$.
Wow, it just clicked for me. Thank you so much for your help.