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I have this mathematical induction problem $$\sum_{i=0}^n j!j = (n + 1)! - 1$$

I want to show that

$$\sum_{i=0}^{k+1} j!j +((k+1)!(k+1)) = (k + 2)! - 1$$

My steps after this line would be to sub the LHS with the original formula:

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

And after factorizing out $(k+1)!$, im left with:

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

Where did I go wrong? Can I not factor out $(k+1)!$ this way?

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  • $\begingroup$ Why are you trying to show that exactly? I may be interpreting this wrong, but I think for your induction step, you want to show that $$\sum_{j=0}^{k+1} j!j = (k + 2)! - 1$$ with the assumption that $$\sum_{j=0}^k j!j = (k+1)! - 1$$. So you'd want to show that $$\sum_{j=0}^k j!j + (k+1)!(k+1) = (k+2)! - 1$$ $\endgroup$ Apr 18 '17 at 15:43
  • $\begingroup$ @OsamaGhani: That's exactly what the OP was asking for. You just repeated the question. $\endgroup$
    – sranthrop
    Apr 18 '17 at 15:44
  • $\begingroup$ @sranthrop the OP's indexing on the induction step was wrong, which led to them simplifying the wrong expression $\endgroup$ Apr 18 '17 at 17:36
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You want to show that

$$ \sum_{i = 0}^k i!i + (k + 1)!(k + 1) = (k + 2)! - 1. $$

Notice the limits on the summation.

This gives you

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

Can you spot where you made the error?

Edit:

Perhaps it will help to let $a = (k + 1)!$. Then you have

$$ \begin{align*} a - 1 + a(k + 1) &= a(1) - 1 + a(k + 1) \\ &= a(1) + a(k + 1) - 1 \\ &= a(1 + (k + 1)) - 1 \\ \end{align*} $$

Notice that the third term does not have an "$a$" in front of it, so we leave it alone when factoring out $a$.

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  • $\begingroup$ I still can't get it =[, if i take out (k+1)! from the equation, wouldn't i get (k+1)! [(1)-1+(1)(k+1)] = (k+1)! [(0)+(k+1)]? $\endgroup$ Apr 18 '17 at 15:53
  • $\begingroup$ @BloopieBloops I've edited the answer. Does it help you understand it better? $\endgroup$ Apr 18 '17 at 15:57
  • $\begingroup$ yes! Subbing it with a finally helped me see, too much 1s got me all confused, thanks! $\endgroup$ Apr 18 '17 at 16:00
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This is the induction step: $$(k+1)!-1+(k+1)!(k+1) = (k+1)!(1+k+1)-1 \\= (k+1)!(k+2)-1 = (k+2)!-1$$

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Base case

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

Inductive hypothesis

Suppose $\sum_{i=0}^{k} j!j = (k+1)! - 1$

We must show that when the inductive hypothesis holds

$\sum_\limits{i=0}^{k+1} j!j = (k + 2)! - 1\\ \sum_\limits{i=0}^{k} j!j + (k+1)!k \\ (k+1)!-1 + (k+1)!k $

By the inductive hypothesis

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

QED

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Let be $ \sum_{i=0}^{k} j! j = (k+1)! -1$ (*) your induction assumption.

So you have then $ \sum_{i=0}^{k+1} j! j = \sum_{i=0}^{k} j! j + (k+1)!(k+1) = (k+1)! -1 +(k+1)!(k+1) = (k+2)! -1$

In second equation I use (*).

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