Binomial with 3 terms inside brackets

**terrorsquid** · November 11th 2011, 10:58 PM

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

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

$2i = 4$

$i = 2$

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

$= 10$

**veileen** · November 12th 2011, 12:20 AM

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

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

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 ^^

**Opalg** · November 12th 2011, 12:21 AM

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 $x^4.$

**HallsofIvy** · November 12th 2011, 11:20 AM

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

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