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First of all, is it called "convert" in English or there is a proper word for this purpose?

I am struggling with the algebra, please help me convert:

$(k+1)!\times(k+2)−1$ to $(k+2)!−1$

And what is the rule for this step?

Best wishes,

  • 2
    ...the definition of factorial states that $(n+1)!=n!(n+1)$2012-04-03
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    You remember the definition of factorial.2012-04-03
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    "convert" is OK, if a bit unusual in this context. Instead of "convert $P$ to $Q$" I'd probably say "show $P$ is equal to $Q$".2012-04-03
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    How to convert 1+1 in 2 ?2012-04-03
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    @Student: [That proof](http://en.wikipedia.org/wiki/File:Principia_Mathematica_theorem_54-43.png) takes some 300 pages or so...2012-04-03
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    @ The Chaz : Well, depends on your choice for axiomatic...2012-04-03
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    Follow from [here](http://math.stackexchange.com/q/127504/19341)...2012-04-03

3 Answers 3

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$$\rm \color{Blue}{(k+1)!}\cdot \color{Green}{(k+2)}=\color{Blue}{1\cdot2\cdot3\cdots k\cdot(k+1)}\cdot \color{Green}{(k+2)}=(k+2)!$$

There's no special "rule" at play above; if you know the definition of the factorial this is immediate.

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    Nice touch with the colors. I didnt know you could do that!2012-04-03
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    At least something useful from this post.2012-04-03
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Without need, $k$ was assumed to be an integer here. One can instead use this definition $$ z!=\Gamma(z+1)=\int_0^\infty t^{z} e^{-t}\, \mathrm{d}t. \! $$ which converges absolutely, if the real part of the complex number z is positive $(\Re(z) > 0)$, to show that $$ (k+2)!=\Gamma(k+3)=\int_0^\infty t^{k+2} e^{-t}\, \mathrm{d}t=\underbrace{\left[-t^{k+2}e^{-t}\right]_0^\infty}_{=0}+(k+2)\int_0^\infty t^{k+1} e^{-t}\, \mathrm{d}t =(k+2)\Gamma(k+2)=(k+2)\cdot (k+1)! $$

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For any positive integer $n$, $n!$ is defined as, $$n!=n(n-1)\cdots1.$$

From the above definition it follows that $n!=n(n-1)!,\forall n\in\mathbb{Z^+}.$

Hence it follows that $(k+1)!\times(k+2)−1=(k+2)!−1.$