Can someone help me understand the trinomial expansion?

[rock.freak667]

Well according to what I've read

$$(a+b+c)^n=\sum_{i,j,k} \left(<br /> \begin{array}{c}<br /> n\\<br /> i,j,k<br /> \end{array}<br /> \right)a^i b^j c^k$$



$$\left(<br /> \begin{array}{c}<br /> n\\<br /> i,j,k<br /> \end{array}<br /> \right)<br /> =\frac{n!}{i!j!k!}$$

I understand the last equation but how would I find the values for i,j and k?

for example if I have $(1+x+x^2)^8$ how would I find the coefficient of x^3 without expanding the entire thing out?

 

[Ben Niehoff]

This:

$$\sum_{i,j,k}$$

means to sum over all (i,j,k) from 0 to n such that i+j+k = n.

 

[slider142]

In other words, find (i,j,k) such that (1^i)*(x^j)*(x^2)^k = x^3 and i+j+k = 8. Then just evaluate that particular multinomial coefficient.

 

[rock.freak667]

so from:(1^i)*(x^j)*(x^2)^k = x^3
j+2k=3 and i+j+k=8...but since I don't have a 3rd equation how would i find those specific values for i,j and k ?

 

[Ben Niehoff]

There are two distinct possibilities. You will have to find both trinomial coefficients and add them together.

Hint: You know j and k are integers between 0 and 8. Just try stuff.

 

[rock.freak667]

Well the only way j+2k=3 would be if j=1,k=1 and so i=6?

then the coefficient of x^3 would be 56?

 

[Ben Niehoff]

Remember that j and/or k can also be zero, so there is one additional possibility.

You're very close, though. :)

 

[rock.freak667]

Well after picking k=0 I got the coefficient to now be 112 and i think that is all i can guess j and k to be.

 

[mirsamir]

Find procedure to solve trinomial expansions:
29 Mar 2009 ... Mir's trinomial expansion theorem - download at 4shared. Mir's trinomial expansion theorem is hosted at free file sharing service 4shared.
www.4shared.com/document/.../Mirs__trinomial_expansion_theo.html - Cached

 

[HallsofIvy]

With a= 1, b= x, $c= x^2$, $a^ib^jc^k$ is equal to $1^ix^j(x^2)^k= x^{j+2k}= x^8$ so j+ 2k= 8 as said before. If k= 0, j= 8. If k= 1, j= 6. If k= 2, j= 4. If k= 3, j= 2. If k= 4, j= 0. The condition that i+ j+ k= 8 gives i= 0, 1, 2, 3, and 4 respectively.

Therefore, the coefficient of $x^8$ is the sum of
$$\begin{pmatrix}8 \\ 0, 8, 0\end{pmatrix}+ \begin{pmatrix}8 \\ 1, 6, 1\end{pmatrix}+ \begin{pmatrix}8 \\ 2, 4, 2\end{pmatrix}+ \begin{pmatrix}8 \\ 3, 2, 3\end{pmatrix}+ \begin{pmatrix}8 \\ 4, 0, 4\end{pmatrix}$$