# Thread: [SOLVED] Need some help with an induction proof

1. ## [SOLVED] Need some help with an induction proof

I'm having some issues with the inductive step for this problem, and I could use a hand with it.

Let $\displaystyle F(n)$ be the statement that $\displaystyle 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1$ whenever $\displaystyle n$ is a positive integer.

I have the basis step and inductive hypothesis done, but I'm having some trouble with the induction proof for $\displaystyle F(k+1)$. Does anyone have any advice?

2. Originally Posted by Runty
I'm having some issues with the inductive step for this problem, and I could use a hand with it.

Let $\displaystyle F(n)$ be the statement that $\displaystyle 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1$ whenever $\displaystyle n$ is a positive integer.

I have the basis step and inductive hypothesis done, but I'm having some trouble with the induction proof for $\displaystyle F(k+1)$. Does anyone have any advice?
Hi Runty,

Sure!

F(n)

$\displaystyle (1)1!+(2)2!+(3)3!+.....+(n)n!=(n+1)!-1$ ??? (1)

If this is so, then the following must be true because of that

$\displaystyle (1)1!+(2)2!+(3)3!+....(n)n!+(n+1)(n+1)!=(n+2)!-1$

To discover whether or not this is so, we utilise statement (1)

F(k+1)

$\displaystyle (1)1!+(2)2!+....(k)k!+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!$

$\displaystyle =(k+1)!(1+k+1)-1$

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

3. Originally Posted by Runty
I'm having some issues with the inductive step for this problem, and I could use a hand with it.

Let $\displaystyle F(n)$ be the statement that $\displaystyle 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1$ whenever $\displaystyle n$ is a positive integer.
Say that we know that $\displaystyle \sum\limits_{k = 1}^N {k \cdot k!} = \left( {N + 1} \right)! - 1$
Then
$\displaystyle \begin{array}{rcl} {\sum\limits_{k = 1}^{N + 1} {k \cdot k!} } & = & {\sum\limits_{k = 1}^N {k \cdot k!} + \left( {N + 1} \right) \cdot \left( {N + 1} \right)!} \\ {} & = & {\left( {N + 1} \right)! - 1 + \left( {N + 1} \right) \cdot \left( {N + 1} \right)!} \\ {} & = & {\left( {N + 1} \right)! \cdot (N + 2) - 1} \\ \end{array}$