How would you prove that via induction? $\sum_{k=1}^{n}k! \cdot k= (n+1)! -1$

Question

Task from old exam:

Prove that $$\sum_{k=1}^{n}k! \cdot k= (n+1)! -1$$

I'd like to use induction proof and nothing else.

Here is what I did:

$n=1:$ $\sum_{k=1}^{1}1! \cdot 1= (1+1)! -1 \Leftrightarrow 1=1$

So this is true, so it will work for an $n$ already.

It worked for one $n$, now we need to show it will work for every other $n$, too.

$$\sum_{k=1}^{n+1}k! \cdot k= (n+2)! -1$$

But what can I do from this point on? There is this sum symbol on the left and I don't know how to continue with it. Any hints maybe?

I really don't want know another way of solving it. It might be easier with telescope or something else but the recommendation was induction.

Answer 1

\begin{align}\sum_{k = 1}^{n+1} k!k &= \sum_{k = 1}^n k!k + (n+1)!(n+1) \overset{*}{=} (n+1)! - 1 + (n+1)!(n+1) \\ &= (n+1)![1 + (n+1)]-1 = \cdots\end{align}

Take it from there. Note the induction hypothesis was used at (*).

Answer 2

\begin{align*}\sum_{k=1}^{n+1} k! \cdot k &= \sum_{k=1}^n k! \cdot k + (n+1)!\cdot (n+1) \\ & = (n+1)! - 1 + (n+1)! \cdot \color{blue}{(n+1)} \\ & = (n+1)! (1 + \color{blue}{n+ 1}) - 1 \\ & = (n+2)(n+1)! - 1 \\ & = (n+2)! - 1 \end{align*}

Answer 3

Note that: $$ \begin{gathered} S(n) = \sum\limits_{k = 1}^n {k!k} = \sum\limits_{k = 1}^n {k!\left( {k + 1 - 1} \right)} = \hfill \\ = \sum\limits_{k = 1}^n {\left( {k + 1} \right)!} - \sum\limits_{k = 1}^n {k!} \hfill \\ \end{gathered} $$ then you can for sure continue

Answer 4

Nearly always if you need to prove $\sum_{i=0}^n gingkabox(i) = floopadoop(n)$ you nearly always notice:

$\sum_{i=0}^{n+1} gingkabox(i) = floopadoop(n) + gingkabox(n+1)$

and the whole thing just becomes a matter of proving

$floopadoop(n) + gingkabox(n+1) = floopadoop(n+1)$.

So can you prove $[(n+1)! - 1] + [(n+1)!(n+1)] = (n+2)! - 1$?

That's all you have to do.

[

$[(n+1)! - 1] + [(n+1)!(n+1)] = $

$[(n+1)! + (n+1)!(n+1)] - 1=$

$(n+1)![1 + (n+1)] -1=$

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

$(n+2)! - 1$

]