0
$\begingroup$

I am stuck halfway while using mathematical induction to prove that $\left(r+1\right)!-r!=r\left(r!\right)$

I know that the first step is to prove the basis is true, therefore, I substituted r=1 into the equation, proving $1$ = $1$ therefore $LHS$ = $RHS$.

The second step is to assume $\left(r+1\right)!-r!=r\left(r!\right)$ is true for $r=k$ therefore, $\left(k+1\right)!-k!=k\left(k!\right)$ and thus, I need to use this assumption to prove that $k+1$ is true. But I do not know how to prove $k+1$ is true. When I substitued $r=k+1$, I got $\left(k+2\right)!-(k+1)!=(k+1)\left((k+1\right))!$.

How do i proceed from here?

$\endgroup$
  • 3
    $\begingroup$ Why do you need induction? $$ (r+1)!-r! = (r+1)r!-r! = ((r+1)-1)r! = r (r!) $$ $\endgroup$ – MisterRiemann Sep 2 '18 at 8:11
  • $\begingroup$ I was told to use the induction method to prove to ensure that we know the concept of mathematical induction. I showed the above mentioned method but I was told even though I proved it, it was not the method wanted $\endgroup$ – deviljones Sep 2 '18 at 8:14
  • 2
    $\begingroup$ You can do what Sobi did with $(r + 1)! - r!$ for $r = k + 1$ to complete the induction step. $\endgroup$ – N. F. Taussig Sep 2 '18 at 8:16
  • $\begingroup$ Don't understand why I didnt think of it... thanks! $\endgroup$ – deviljones Sep 2 '18 at 8:38
0
$\begingroup$

$$\begin{array}{l}\left(r+1\right)!-r!=\left(r+1\right)r!-r!=\left(\left(r+1\right)-1\right)r!=r\left(r!\right)\\\end{array}$$

by substituting $$r=k+1$$

$$ \begin{array}{l}\left(k+1+1\right)!-\left(k+1\right)!\;\\=\;\left(k+2\right)!-\left(k+1\right)!\;\\=\left(k+2\right)\left(k+1\right)!-\left(k+1\right)!\;\\=\left(k+1\right)!\left(k+2-1\right)\\=\left(k+1\right)!\left(k+1\right)\end{array}$$ $$\begin{array}{l}=\left(k+1\right)\left(k+1\right)!\\=r\left(r!\right)\end{array}$$

$\endgroup$
0
$\begingroup$

As noticed in the comments we don't need induction since we have

$$\left(r+1\right)!-r!=r!(r+1-1)=r\left(r!\right)$$

and the given identity holds.

$\endgroup$
  • $\begingroup$ Yes we don't have to, but I was explicitly told to use the induction method to prove instead. Nevertheless, thanks! $\endgroup$ – deviljones Sep 2 '18 at 8:43
  • $\begingroup$ @deviljones Yes we can of course but the induction step reduces to the same identity which is trivial. $\endgroup$ – gimusi Sep 2 '18 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.