Mathematical Induction Proof: $\sum_{i=1}^n$ $(i \times i!) = (n+1)! - 1$

Question

I am to use mathematical induction to prove:

$\sum_{i=1}^n$ $(i \times i!) = (n+1)! - 1$

my base case is n = 1

$RHS: (1 \times1!) = 1$

$LHS: (1+1)! - 1 = 1$

If I am not mistaken the next step is to assume n=k so

$=(k+1)! -1$

then I need to show k+1

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

now I'm a bit confused on where to go from here. I THINK my next step is:

$(k+1)!-1 + (k+1 \times k+1!)$

Is this the correct next step? if it is how do I algebraically show that it is equal to $(k+2)!-1$

Any help is appreciated guys! Thanks!

Answer 1

The next is indeed $$(k+1)!-1+(k+1)(k+1)!=(k+1+1)(k+1)!-1=(k+2)!-1.$$

This remarkable identity can be used to define a factorial basis of the integers.

Answer 2

So let's assume that $$ \sum_{i=1}^k (i\cdot i!)=(k+1)!-1. $$ Then for the $k+1$ case, we have $$ \sum_{i=1}^{k+1}(i\cdot i!)=(k+1)\cdot (k+1)!+\sum_{i=1}^k(i\cdot i!)=(k+1)\cdot (k+1)!+(k+1)!-1. $$ Factoring out $(k+1)!$ from the last term, we have $$ \sum_{i=1}^{k+1}(i\cdot i!)=(k+1)![1+k+1]-1=(k+2)!-1 $$ just like you want.

Answer 3

You already verified the base case.

If it holds for $n$, let us demonstrate it for $n+1$. We know that: $$ \sum_{i=1}^nii!=(n+1)!-1, $$ and $$ \sum_{i=1}^{n+1}ii!=\sum_{i=1}^nii!+(n+1)(n+1)!. $$ The first term, $\sum_{i=1}^nii!=(n+1)!-1,$ is, for induction assumption, equal to $(n+1)!-1$, thus: $$ \sum_{i=1}^nii!+(n+1)(n+1)!=(n+1)!-1+(n+1)(n+1)!=(n+1)!(1+n+1)-1= $$ $$ =(n+2)!-1, $$ Hence: $$ \sum_{i=1}^{n+1}ii!=(n+2)!-1, $$ q.e.d.