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(
\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?

[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 dont 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.