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Prove this by induction: $$\sum_{i=1}^n i(i!) = (n+1)!-1$$

So I wrote:

Base Case: $n=1$ so $1(1!) = 1$ and $(1+1)!-1 = 1$.

Let $n=k$ so that $$\sum_{i=1}^ki(i!)=(k+1)!-1$$

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

But I'm stuck here. Also is there any easier way I can write out summation on stack exchange? A format for writing it out on one line? Or is my way preferred?

M47145
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Adam
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  • You can use common latex syntax to write mathematical formulas. As an example $\sum_1^n i\cdot i!$ is how I wrote the sum. You can read more here: http://math.stackexchange.com/editing-help – Jared Apr 07 '14 at 04:32
  • $\sum_1^n i\cdot i!$ I see – Adam Apr 07 '14 at 04:34
  • Also, you can click edit on any answer or question to view their source (just make sure you discard your edit if you do that). – Jared Apr 07 '14 at 04:35

1 Answers1

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$$ \sum_1^n i\cdot i! = \sum_1^{n - 1} i\cdot i! + n\cdot n! $$

Now plug in the formula for $\sum i\cdot i!$ for $n - 1$, add the next term and see if you get the correct formula for $n$:

$$ ((n - 1) + 1)! - 1 + n\cdot n! = n! - 1 + n\cdot n! = (n + 1)n! - 1 $$

That's the inductive step. If the formula holds for $n - 1$ then it also holds for $n$.

Jared
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  • So when dealing with factorials we use k-1 not k+1? – Adam Apr 07 '14 at 04:30
  • Also how did you make your notation so neat? Did you use alt-w and make it large font? Haha I'm just trying to figure out the best way. – Adam Apr 07 '14 at 04:31
  • You can do it either way. You can use $n$ then prove the formula is still correct for $n + 1$. I prefer going from $n - 1$ to $n$ because the final formula will be the exact formula (i.e. for $n$). If you start with $n$, then the final result should be $(n + 2)! - 1$. – Jared Apr 07 '14 at 04:31
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    @adam - To see how a user did something, just $\color{grey}{edit}$ the question. Make sure to $\color{grey}{cancel}$ to get out. You can also right-click on a math equation and select Show Math As/TeX Commands, then paste into an answer box. Erase all of the text before you leave that page. – Steven Alexis Gregory Oct 28 '16 at 14:33