[SOLVED] Need some help with an induction proof

**Runty** · February 25th 2010, 06:42 AM

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

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

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

**Archie Meade** · February 25th 2010, 07:20 AM

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

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

Hi Runty,

Sure!

F(n)

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

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

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

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

F(k+1)

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

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

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

**Plato** · February 25th 2010, 07:28 AM

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

Say that we know that $\sum\limits_{k = 1}^N {k \cdot k!} = \left( {N + 1} \right)! - 1$
Then
$ \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}$

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