Jason Farmer

2021-11-04

Prove that $1·1!+2·2!+···+n·n!=\left(n+1\right)!-1$ whenever n is a positive integer.

i1ziZ

To proof $1\cdot 1!+2\cdot 2!+\dots +n\cdot n\ne \left(n+1\right)!-1$ for every positive integer n.
Proof by induction
Let $P\left(n\right)$ be $1\cdot 1!+2\cdot 2!+\dots +n\cdot n\ne 1\cdot 1=\left(n+1\right)!-1$
Basis step n=1
$1\cdot 1!+2\cdot 2!+\dots +n\cdot n\ne 1\cdot 1\ne 1\cdot 1=1$
$\left(n+1\right)!-1=\left(1+1\right)!-1=2!-1=2-1=1$
We then note P(1) is true.
Induction step let $P\left(k+1\right)$ is also true
$1\cdot 1!+2\cdot 2+\dots +k\cdot k\ne \left(k+1\right)!-1$
We need to prove that $P\left(k+1\right)$ is also true.
$1\cdot 1!+2\cdot 2!+\dots +k\cdot k!+\left(k+1\right)\cdot \left(k+1\right)!$
$=\left(k+1\right)!-1+\left(k+1\right)\cdot \left(k+1\right)!$
$=1\cdot \left(k\cdot 1\right)!-1+\left(k+1\right)\cdot \left(k+1\right)!$
$=1\cdot \left(k+1\right)!+\left(k+1\right)\cdot \left(k+1\right)!-1$
$=\left(1+k+1\right)\left(k+1\right)!-1$
$=\left(k+2\right)!-1$
$=\left(\left(k+1\right)+1\right)!-1$
We then note that $P\left(k+1\right)$ is also true

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