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FritoTaco

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## Homework Statement

Prove that [itex]1\cdot1!+2\cdot2!+...+n\cdot n! = (n+1)!-1[/itex] whenever [itex]n[/itex] is a positive integer.

## Homework Equations

## The Attempt at a Solution

I'm having trouble simplifying towards the end of the proof.

__Proof:__

Let [itex]P(n)[/itex] be the statement [itex]1\cdot1!+2\cdot2!+...+n\cdot n! = (n+1)!-1[/itex]

__Basis step:__

We want to show [itex]P(1)[/itex] is true, where n = 1. Show both sides of the equation to be true (i.e. Left-hand side and right-hand side equal to each other)

LHS

[itex]n\cdot n![/itex]

[itex]= 1\cdot1![/itex]

[itex]= 1\cdot1[/itex]

[itex]= 1[/itex]

RHS

[itex](n+1)!-1[/itex]

[itex]= (1+1)! - 1[/itex]

[itex]= 2! - 1[/itex]

[itex]= 1[/itex]

Both sides are equal.

__Induction step:__

For the inductive hypothesis, assume [itex]P(k)[/itex] is true for some [itex]k\geq1[/itex]; then show [itex]P(k+1)[/itex] is true.

assume [itex]P(k): 1\cdot1!+2\cdot2!+...+k\cdot k! = (k+1)!-1[/itex] (

*we replaced [itex]n[/itex] with [itex]k[/itex] from original equation)*

show [itex]P(k+1): (1\cdot1!+2\cdot2!+...+k\cdot k!) + (k+1)(k+1)! = (k+1+1)!-1 + (k+1)(k+1)![/itex]

on the left side on the equation, we kept [itex]k\cdot k![/itex] but also added and replaced the [itex]k[/itex] in [itex]k\cdot k![/itex] with [itex]k+1[/itex]

on the right side of the equation we added [itex](k+1)(k+1)![/itex] that was also on the left side (so we essentially added [itex](k+1)(k+1)![/itex] on both sides of the equation.

Simplify right side of equation:

[itex]= (k+2)!-1+(k+1)(k+1)![/itex]

We want the right-hand side of [itex]P(k)[/itex] which is [itex](k+1)!-1[/itex] to look like [itex](k+2)!-1[/itex].

Okay, so this is where I'm having trouble getting the equation (which is [itex]P(k): (k+1)! -1[/itex]) to look like my [itex]P(k+1)[/itex] which is [itex](k+2)!-1[/itex].

[itex]=[(k+1)!-1]+(k+1)(k+1)![/itex]

we already have [itex]-1[/itex] in the equation so I feel like we just need to manipulate the rest.

I don't know how to condense [itex](k+1)! +(k+1)(k+1)![/itex]

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[itex]=(k+2)!-1[/itex]