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Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.

I started, but I don't understand how to continue

$(1 + x + x^2 + x^3 ...)^2 \cdot (1 + x^2 + x^4 + x^6 ...)$

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

$=\big(1 + \binom{1 + 2 - 1}{1}x + \binom{2 + 2 - 1}{2}x^2 ...\big) \cdot ???$

How do you find coefficient to $x^{10}$?

Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $\frac{1}{1-x^2}$ how?

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Notice that $\left(\frac{1}{1-x}\right)^2=\frac{d}{dx}\left(\frac{1}{1-x}\right)$, and hence $\left(\frac{1}{1-x}\right)^2=1+2x+3x^2+\dots$ You can then find the $x^{10}$ coefficient just by multiplying out.

The equality stated in the textbooks holds for the same reason that $1+x+x^2+\dots=\frac{1}{1-x}$.

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The original problem is $(1 + x + x^2 + x^3 ...)^2 \cdot (1 + x^2 + x^4 + x^6 ...)$

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

Which can be shown by,

$1 = (1 + x + x^2 + x^3 ...)(1-x)$

$1 = (1 + x + x^2 + x^3 ...) - x \cdot (1 + x + x^2 + x^3 ...)$

$1 =1$

To show, $\frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$

we substitute $y = x^2$

$\frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $\frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.

To solve the equation,

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

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

$=\big(1 + \binom{1 + 2 - 1}{1}x + \binom{2 + 2 - 1}{2}x^2 ...\big) \cdot (1 + x^2 + x^4 + x^6 ...)$

Thus the coefficient of $x^{10}$ is the coefficient of

$=1 \cdot x^{10} + \binom{3}{1}x^2 \cdot x^{8} + \binom{5}{1}x^4 \cdot x^{6} + \binom{7}{1}x^6 \cdot x^{4} + \binom{9}{1}x^8 \cdot x^{2} + \binom{11}{1}x^{10} \cdot 1$

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