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I want to prove with mathematical induction that:

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

So in the first step we define $n = 1$:

$$\sum_{i=1}^1 i \cdot i! = 1 \cdot 1! = 1 = 2! - 1$$

In the second step we define $n = n + 1$:

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

This can now be simplified to $(n+2)(n+1)! - 1$ but I do not see how...

We could do $(n+1)!(-1 + (n+1)) = (n+1)! \cdot n$ but still this will not be the same as above.

What did I miss?

Bodo

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You are almost there.

Note that $$ (n+1)! - 1 + (n+1)(n+1)! = -1 + (n+1)!( 1 + (n+1) ) = -1+ (n+2)(n+1)! = (n+2)! - 1. $$

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  • $\begingroup$ What "rule" allows the transformation from $-1 + (n+2)(n+1)! = (n+2)! -1$? $\endgroup$ – bodokaiser Jan 3 '15 at 13:17
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    $\begingroup$ The rule $k\times (k-1)! = k!$, applied with $k=n+2$. $\endgroup$ – user133281 Jan 3 '15 at 13:19
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$(n+1)!-1+(n+1)(n+1)!=(n+1)!(n+1+1)-1=(n+1)!(n+2)-1$

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