# Can someone help me understand the trinomial expansion?

1. Dec 13, 2007

### rock.freak667

Well according to what I've read

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

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

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?

2. Dec 13, 2007

### 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.

3. Dec 13, 2007

### 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.

4. Dec 13, 2007

### rock.freak667

so from:(1^i)*(x^j)*(x^2)^k = x^3
j+2k=3 and i+j+k=8...but since I dont have a 3rd equation how would i find those specific values for i,j and k ?

5. Dec 13, 2007

### 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.

6. Dec 13, 2007

### 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?

7. Dec 13, 2007

### Ben Niehoff

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

You're very close, though. :)

8. Dec 13, 2007

### 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.

9. Jun 10, 2011

### mirsamir

10. Jun 11, 2011

### HallsofIvy

Staff Emeritus
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
$$\begin{pmatrix}8 \\ 0, 8, 0\end{pmatrix}+ \begin{pmatrix}8 \\ 1, 6, 1\end{pmatrix}+ \begin{pmatrix}8 \\ 2, 4, 2\end{pmatrix}+ \begin{pmatrix}8 \\ 3, 2, 3\end{pmatrix}+ \begin{pmatrix}8 \\ 4, 0, 4\end{pmatrix}$$

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