Binomial with 3 terms inside brackets

What is the coefficient of $\displaystyle x^4$ in the series expansion of $\displaystyle (1+x+x^2)^{-4}$?

This would be easy if it were just $\displaystyle (1+x^2)^{-4}$;however, I don't know what to do with the three terms.

Normally I would find the coefficient of such a problem as follows:

$\displaystyle x^4$ in $\displaystyle (1+x^2)^{-4}$

= $\displaystyle -4\choose{i}$$\displaystyle (x^2)^i$

$\displaystyle 2i = 4$

$\displaystyle i = 2$

$\displaystyle 5\choose{2}$$\displaystyle (1)^2$

$\displaystyle = 10$

Re: Binomial with 3 terms inside brackets

$\displaystyle (1+x+x^2)^{-4}=\frac{1}{(1+x+x^2)^4}$

$\displaystyle (1+x+x^2)^4=[x^2+(1+x)]^4=\sum_{k=0}^{4}\binom{4}{k}(x^2)^{4-k}(1+x)^{k}=\sum_{k=0}^{4}\left [\binom{4}{k}x^{8-2k}\sum_{i=0}^{k}\binom{k}{i}1^{k-i}x^i\right ]=$$\displaystyle \sum_{k=0}^{4}\left [\binom{4}{k}x^{8-2k}\sum_{i=0}^{k}\binom{k}{i}x^i\right ]$

You need the coefficient of $\displaystyle x^4$, so:

$\displaystyle 8-2k+i=4, \: k=\overline{0,4} \: and \: i=\overline{0,k}$. Find k and i ^^

Re: Binomial with 3 terms inside brackets

Quote:

Originally Posted by

**terrorsquid** What is the coefficient of $\displaystyle x^4$ in the series expansion of $\displaystyle (1+x+x^2)^{-4}$?

This would be easy if it were just $\displaystyle (1+x^2)^{-4}$;however, I don't know what to do with the three terms.

$\displaystyle 1+x+x^2 = \frac{1-x^3}{1-x}$ (sum of geometric series).

Therefore $\displaystyle (1+x+x^2)^{-4} = (1-x)^4(1-x^3)^{-4} = (1-4x+6x^2-4x^3+x^4)(1+4x^3+\ldots)$, from which you can easily pick out the coefficient of $\displaystyle x^4.$

Re: Binomial with 3 terms inside brackets

By the way, **because** there are three terms, this is a "tri"-nomial, not a "bi"-nomial?