0
$\begingroup$

This question already has an answer here:

Assume $n = k$:

$$(k+1)! -1 = 1!\cdot 1!\cdot1+...+k!\cdot k$$

To test for $k + 1$, I add $k+1$ to the sum as new element and equate its sum with that of the first $k$ elements with the formula adjusted for $k+1$:

$$(k+1)!-1+(k+1)!(k+1) = ((k+1)+1)! -1$$ $$(k+1)!+(k+1)!(k+1) = (k+1)(k+1)!$$ $$(k+1)! = 0$$

Equation doesn't hold water. What I did do wrong?

$\endgroup$

marked as duplicate by Aqua, Mike Earnest, FredH, Cesareo, Strants Apr 3 at 21:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ you have commas where you should have put plus signs $\endgroup$ – Will Jagy Apr 3 at 18:48
  • $\begingroup$ @WillJagy Thanks, fixed it. $\endgroup$ – daedsidog Apr 3 at 19:09
  • $\begingroup$ Where does this come from: $(k+1)!-1+(k+1)!(k+1) = ((k+1)+1)! -1$? $\endgroup$ – eyeballfrog Apr 3 at 19:13
  • $\begingroup$ @eyeballfrog Hopefully my edit will clear things up. $\endgroup$ – daedsidog Apr 3 at 19:20
1
$\begingroup$

First addressing your question on equation doesn't hold water.

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

I'm not sure how you arrived at the right hand side of the equation. You should have $(k+2)! = (k+2)(k+1)!$ if you wish to expand it out.

Now to address the way you are doing your induction. Since you are trying to prove that an equation is true, you should start from the left hand side (LHS) and derive the right hand side (RHS) or at least something to that effect.

The proof should look like:

Let $p(n)$ be the statement $1!\cdot 1+ 2!\cdot2+ 3!\cdot3+ … +n!\cdot n = (n+1)!-1$.

$p(0)$ is true because $1!\cdot 1 = (1+1)!-1$.

Assume $p(n)$ is true, then \begin{align*} \text{LHS} &= 1!\cdot 1+ 2!\cdot2+ 3!\cdot3+ … +n!\cdot n + (n+1)!\cdot (n+1)\\ &= (n+1)!-1 + (n+1)!\cdot (n+1)\\ &= (n+2)(n+1)!-1\\ &= (n+2)! - 1\\ &=\text{RHS} \end{align*}

$\endgroup$
  • $\begingroup$ Thanks, the issue was that I used the incorrect expansion. $\endgroup$ – daedsidog Apr 3 at 19:28

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