Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

$(k+1)!*(k+2)−1$ to $(k+2)!−1$

And what is the rule for this step?

Best wishes,

share|improve this question
2  
...the definition of factorial states that $(n+1)!=n!(n+1)$ –  Alex Becker Apr 3 '12 at 3:38
1  
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
1  
@Student: That proof takes some 300 pages or so... –  The Chaz 2.0 Apr 3 '12 at 5:02
show 2 more comments

2 Answers

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

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|improve this answer
add comment

Your Answer

 
discard

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.