Paradoxx said:
For example,if I have ##(a+b)^n## , where n can be a fraction...I know I solve like:
##(a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...##
If I had ##(a-b)^n## I should alternate the signal + and -...
But in the case ##(a+b+c)^n## or more precisely ##(a+b-c)^n## I don't know how to do it...
I just want to expand like the one above...
You have
still have not answered my original question: expand with respect to what?
You could expand in power of ##a## as
[tex](a+b+c)^n = (b+c)^n \sum_{k=0}^{\infty} {n \choose k} \left( \frac{a}{b+c} \right)^k,[/tex]
where
[tex]{n \choose k}= \frac{n(n-1) \cdots (n-k+1)}{k!}[/tex]
for any real ##n## and non-negative integer ##k##.
Or, you could try to expand the whole thing in powers like ##a^k b^j c^l##, perhaps by re-writing the above as
[tex](a+b+c)^n = \sum_{k=0}^{\infty} {n \choose k} a^k (b+c)^{n-k},[/tex]
then expanding each of the ##(c+b)^{n-k}## in powers of ##b## and ##c##, and then collecting terms. It would be messy.
Or, you could use the trinomial expansion, which says that
[tex](a+b+c)^n = \sum_{k_1,k_2,k_3} c(k_1,k_2,k_3) a^{k_1} b^{k_2} c^{k_3},[/tex]
where
[tex]c(k_1,k_2,k_3) = {n \choose k_1} {n-k_1 \choose k_2} {n-k_1-k_2 \choose k_3}[/tex]