# Need help proving mathematical induction

• Aug 28th 2009, 03:04 PM
kalel918
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,
\$\displaystyle n(n!) = (n+1)*(n+1)!\$

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

This is where I am stuck. What next?
• Aug 28th 2009, 03:12 PM
mr fantastic
Quote:

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

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

This is where I am stuck. What next?

\$\displaystyle (n+1)! - 1 + (n+1)*(n+1)! = (n+1)! (1 + [n+1]) - 1 = (n+1)!(n+2) - 1 = (n+2)! - 1\$
• Aug 28th 2009, 03:18 PM
kalel918
Quote:

Originally Posted by mr fantastic
\$\displaystyle (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.

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

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

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

Okay, you lost me here. Can you please clarify this once more?
• Aug 28th 2009, 05:49 PM
mr fantastic
Quote:

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.

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

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

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

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

\$\displaystyle (n + 2)! = (n+2) {\color{red}(n+1)n(n-1) ..... 1} = (n+2) {\color{red}(n+1)!}\$.
• Aug 29th 2009, 08:16 AM
kalel918
Quote:

Originally Posted by mr fantastic
\$\displaystyle (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.