# Binomial with 3 terms inside brackets

• November 11th 2011, 10:58 PM
terrorsquid
Binomial with 3 terms inside brackets
What is the coefficient of $x^4$ in the series expansion of $(1+x+x^2)^{-4}$?

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

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

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

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

$2i = 4$

$i = 2$

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

$= 10$
• November 12th 2011, 12:20 AM
veileen
Re: Binomial with 3 terms inside brackets
$(1+x+x^2)^{-4}=\frac{1}{(1+x+x^2)^4}$

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

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

$8-2k+i=4, \: k=\overline{0,4} \: and \: i=\overline{0,k}$. Find k and i ^^
• November 12th 2011, 12:21 AM
Opalg
Re: Binomial with 3 terms inside brackets
Quote:

Originally Posted by terrorsquid
What is the coefficient of $x^4$ in the series expansion of $(1+x+x^2)^{-4}$?

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

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

Therefore $(1+x+x^2)^{-4} = (1-x)^4(1-x^3)^{-4} = (1-4x+6x^2-4x^3+x^4)(1+4x^3+\ldots)$, from which you can easily pick out the coefficient of $x^4.$
• November 12th 2011, 11:20 AM
HallsofIvy
Re: Binomial with 3 terms inside brackets
By the way, because there are three terms, this is a "tri"-nomial, not a "bi"-nomial?