Prove that $1 \cdot 1!+2 \cdot 2!+\cdots+n \cdot n!=(n+1)!-1$

Question

Prove that $1 \cdot 1!+2 \cdot 2!+\cdots+n \cdot n!=(n+1)!-1$ whenever $n$ is a positive integer.

Basis step:

$P(1)$ is true because $1 \cdot 1!=(1+1)!-1$ evaluate to $1$ on both sides.

Inductive step:

We assume that $1 \cdot 1!+2 \cdot 2!+\cdots+k \cdot k!=(k+1)!-1$ for some positive integer $k$.

So under this assumption, it must be shown that $P(k+1)$ is true.

$$1 \cdot 1!+2 \cdot 2!+\cdots+k \cdot k!+(k+1) \cdot (k+1)!=(k+1)!-1+(k+1) \cdot (k+1)!$$

then we have that

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

My question is how my teacher got the last step?

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

Answer 1

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

Answer 2

Well we know that $(k+2)! = (k+2)\times(k+1)\times k \times (k-1) ... \times 1$.

However, we know that this can be expressed as $(k+2)(k+1)!$

And so you don't even need to do anything else. You are basically done.

Here is the general formula.

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

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

and so on

Answer 3

First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}k\cdot{k!}=(1+1)!-1$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}k\cdot{k!}=(n+1)!-1$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}k\cdot{k!}=$

$\color{red}{\sum\limits_{k=1}^{n}k\cdot{k!}}+(n+1)\cdot(n+1)!=$

$\color{red}{(n+1)!-1}+(n+1)\cdot(n+1)!=$

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

$(n+2)!-1$

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Please note that the assumption is used only in the part marked red.