problems with (!) function

**Reminisce** · July 20th 2010, 11:13 PM

Im not sure what ! is called but my question is how is

(x+1)! / x! = (x+1)

also is there a formula to solve

(x+1)^a where a equals a # (for example 18)

**Sudharaka** · July 20th 2010, 11:35 PM

Originally Posted by Reminisce

Im not sure what ! is called but my question is how is

(x+1)! / x! = (x+1)

Dear Reminisce,

$\frac{(x+1)!}{x!}=\frac{(x+1) \times (x) \times (x-1) \times ...... \times 2 \times 1}{(x) \times (x-1) ..... \times 2 \times 1} =x+1$

Did you understand now?

**yeKciM** · July 20th 2010, 11:52 PM

Originally Posted by Reminisce

Im not sure what ! is called but my question is how is

(x+1)! / x! = (x+1)

$5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$
or
$5!=5 \cdot4 !$
$n!=n \cdot (n-1) \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot (n-m)!$
so u have...
$\frac{(x+1)!}{x!}=\frac{(x+1) \cdot x!}{x!}=(x+1)$

**Mathelogician** · July 20th 2010, 11:56 PM

Infact it's called Factorial.

**chisigma** · July 21st 2010, 12:18 AM

Originally Posted by Reminisce

Im not sure what ! is called but my question is how is

(x+1)! / x! = (x+1)

also is there a formula to solve

(x+1)^a where a equals a # (for example 18)

In general is x is a real number with x>-1 [so that it isn't neccessarly an integer...] the factorial function [or 'Pi function' according to Gauss interpretation...] is defined as a definite integral that I don't report because the details are a little complex. If You want more details about it see...

Gamma function - Wikipedia, the free encyclopedia

In particular the 'facrorial function' satisfies the relation...

$x! = x\ (x-1)!$ (1)

Till now we are 'all right'... a little less clear for Me is the second part of Your question, regarding the expression $(x+1)^{a}$ ... could You supply more details please?...

Kind regards

$\chi$ $\sigma$

**Reminisce** · July 21st 2010, 05:00 AM

Thank you very much
as for the second part i was wondering if there is a formula to FOIL anything bigger then 3
example
(x+1)^18
because it would be a hassle to write (x+1) eighteen times and FOILing it like that

**yeKciM** · July 21st 2010, 05:38 AM

u can use : (sorry i do'nt know how this LaTex work just jet)

_____1_____
___1_2_1___
__1_3_3_1__
_1_4_6_4_1_
...................

for let's say 10th... it would be... 1_10_45_120_210_252_210_120_45_10_1

and u'll have
$(x+1)^{10}=x^{10}+10x^{9}+45x^{8}+120x^{7}+210x^{6 }+252x^{5}+210x^{4}+120x^{3}+45x^{2}+10x^{1}+1$

and so on.... but if u need to know let's sey n-th member of that u can use binomial template (or something... I'm bad with English... more or less)

**Reminisce** · July 21st 2010, 06:25 AM

I'm still a bit confused. Like how did you get the values 1_10_45_120....
from using the pyramid

**yeKciM** · July 21st 2010, 07:10 AM

well... when u do pyramid ... see that at each side are ones... but numbers inside are sum od two numbers from above...
look at 4... firs is one ... second is (above 1+3 =4) 4 .... 3rd is (above 3+3 =6) 6 ... 4th is (above 3+1=4) 4 and 5th is 1
and so on...

or like this :

$(a+b)^n = a^n + \binom {n}{1}a^{n-1} b + \binom {n}{2}a^{n-2}b^2 + \cdot \cdot \cdot + \binom{n}{n-1}a b^{n-1}+b^n$

or if u need let's say (k+1)th member u'll use this one :

$T_{(k+1)}=\binom {n}{k}a^{n-k}b^k$ that's one if u need (k+1)th member from begining ...

$T_{(k+1)}=\binom {n}{k}a^{k}b^{n-k}$ that's one if u need (k+1)th member from end ...

**undefined** · July 21st 2010, 07:19 AM

Originally Posted by Reminisce

I'm still a bit confused. Like how did you get the values 1_10_45_120....
from using the pyramid

Pascal's triangle - Wikipedia, the free encyclopedia

Also for doing binomial coefficients you might find this useful

Combinations and Permutations Calculator

Choose No - No from dropdown menus, then for example n = 18 and r = 5 will give you the coefficient in front of the x^5 term of (x+1)^18 expanded. See

Binomial coefficient - Wikipedia, the free encyclopedia

**Unknown008** · July 21st 2010, 07:27 AM

If you want to expand (x+1)^a, then you get:

$(x+1)^a = x^a + \binom{a}{1}(x^{a-1})(1^1) + \binom{a}{2}(x^{a-2})(1^2) + ... + \binom{a}{a}(x^{a-a})(1^a)$

Which is equivalent to:

$(x+1)^a = x^a + \binom{a}{1}(x^{a-1}) + \binom{a}{2}(x^{a-2}) + ... + 1$

As 1 to the power of anything will give 1.

Now, $\binom{a}{1} = \frac{a}{1!}$

$\binom{a}{2} = \frac{a(a-1)}{2!}$

Or, $\binom{a}{n} = \frac{a!}{(a-n)!(n!)}$

If that is what you were looking for...

EDIT: Oops. I didn't see your post yekcim =S

**yeKciM** · July 21st 2010, 08:03 AM

Originally Posted by Unknown008

EDIT: Oops. I didn't see your post yekcim =S

lol it's my fault... it take me too long time to see how to write \binom hehehehe so i didn't see urs

and to make a note that :

$\binom {n}{0} = 1$

$\binom {0}{0} = 1$

$\binom {n}{n} = 1$

$\binom {n}{k} = \binom {n}{n-k}$

**Reminisce** · July 22nd 2010, 02:12 PM

o.O thanks but all those equations confused me but i did however found out how to do them Using Pascal's Triangle, thanks!

This is the way i figured it out:

$(a+b)^n$

$X(a^nb^0)$

X = pascal's number
n = n but as we move to the right we subtract 1
0 = 0 but as we move to the right we add 1

**CaptainBlack** · July 22nd 2010, 02:20 PM

Originally Posted by chisigma

In general is x is a real number with x>-1 [so that it isn't neccessarly an integer...] the factorial function [or 'Pi function' according to Gauss interpretation...] is defined as a definite integral that I don't report because the details are a little complex. If You want more details about it see...

No its not, it is a function defined on the natural numbers. There is a way of extending it as you suggest (to a larger domain as well), but they are different things, and what you suggest here is useless to the poster of a question in the pre-university section of MHF

CB

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