Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,

share|cite|improve this question
...the definition of factorial states that $(n+1)!=n!(n+1)$ – Alex Becker Apr 3 '12 at 3:38
You remember the definition of factorial. – Arturo Magidin Apr 3 '12 at 3:38
"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 $Q$". – Gerry Myerson Apr 3 '12 at 4:23
How to convert 1+1 in 2 ? – Student Apr 3 '12 at 4:42
@Student: That proof takes some 300 pages or so... – The Chaz 2.0 Apr 3 '12 at 5:02
up vote 6 down vote accepted

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

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

share|cite|improve this answer
Nice touch with the colors. I didnt know you could do that! – Daniel Montealegre Apr 3 '12 at 3:43
At least something useful from this post. – Salech Alhasov Apr 3 '12 at 4:09

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 $$ (k+2)!=\Gamma(k+3)=\int_0^\infty t^{k+2} e^{-t}\, \mathrm{d}t=\underbrace{\left[-t^{k+2}e^{-t}\right]_0^\infty}_{=0}+(k+2)\int_0^\infty t^{k+1} e^{-t}\, \mathrm{d}t =(k+2)\Gamma(k+2)=(k+2)\cdot (k+1)! $$

share|cite|improve this answer

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 $(k+1)!\times(k+2)−1=(k+2)!−1.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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