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...
$\displaystyle x! = x\ (x-1)!$ (1)
Till now we are 'all right'... a little less clear for Me is the second part of Your question, regarding the expression $\displaystyle (x+1)^{a}$... could You supply more details please?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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
$\displaystyle (x+1)^{10}=x^{10}+10x^{9}+45x^{8}+120x^{7}+210x^{6 }+252x^{5}+210x^{4}+120x^{3}+45x^{2}+10x^{1}+1$
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)
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 :
$\displaystyle (a+b)^n = a^n + \binom {n}{1}a^{n-1} b + \binom {n}{2}a^{n-2}b^2 + \cdot \cdot \cdot + \binom{n}{n-1}a b^{n-1}+b^n$
or if u need let's say (k+1)th member u'll use this one :
$\displaystyle T_{(k+1)}=\binom {n}{k}a^{n-k}b^k $ that's one if u need (k+1)th member from begining ...
$\displaystyle T_{(k+1)}=\binom {n}{k}a^{k}b^{n-k} $ that's one if u need (k+1)th member from end ...
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
If you want to expand (x+1)^a, then you get:
$\displaystyle (x+1)^a = x^a + \binom{a}{1}(x^{a-1})(1^1) + \binom{a}{2}(x^{a-2})(1^2) + ... + \binom{a}{a}(x^{a-a})(1^a)$
Which is equivalent to:
$\displaystyle (x+1)^a = x^a + \binom{a}{1}(x^{a-1}) + \binom{a}{2}(x^{a-2}) + ... + 1$
As 1 to the power of anything will give 1.
Now, $\displaystyle \binom{a}{1} = \frac{a}{1!}$
$\displaystyle \binom{a}{2} = \frac{a(a-1)}{2!}$
Or, $\displaystyle \binom{a}{n} = \frac{a!}{(a-n)!(n!)}$
If that is what you were looking for...
EDIT: Oops. I didn't see your post yekcim =S
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 :
$\displaystyle \binom {n}{0} = 1$
$\displaystyle \binom {0}{0} = 1$
$\displaystyle \binom {n}{n} = 1$
$\displaystyle \binom {n}{k} = \binom {n}{n-k}$
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:
$\displaystyle (a+b)^n$
$\displaystyle X(a^nb^0)$
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
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