Find the Binomial Series of g(x) = 1/(1+5x^2)

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Find the Binomial Series of g(x) = 1/(1+5x^2)^{4}

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g(x) = (1+5x^{2})^{-}^{4}

g(x) = ∑{k=0, ∞}C(α, k) (5x^{2})^{k}, where α = -4, and

C(α, k) = α(α-1)(α-2)...(α-k+1)/k!

The binomial series is

(1+u)^{n} = ∑_{k=1}^{∞} (n k) u^{k} for |u|<1,

where (n k) is the generalized binomial coefficient, n(n-1)(n-2)...(n-k+1)/k!

Therefore,

(1+5x^{2})^{-4} = ∑_{k=1}^{∞} (-4 k) 5^{k}x^{2k}, where (-4 k) = (-4)(-5)(-6)...(-3-k)/k!

In particular, (-4 1) = -4, (-4 2) = (-4)(-5)/2!=10, (-4 3) = (-4)(-5)(-6)/3! = -20, so that the first three terms are:

(1+5x^{2})^{-4} = -4 (5)x^{2} +10(25)x^{4} -20(125)x^{6} +- ... = -20x^{2} + 250x^{4} - 2500x^{6} +- ...

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