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So Given modified bessel equation (t/a)*I_1(at) using the general formula

I have arrived at the answer.

$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7}+ \cdots$$

And according to wolfram alpha its laurent series is equal to $\frac{1}{(s^2-a^2)^{3/2}}$ equal to my answer above

So the question is how my answer

$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7} + \cdots$$

equal to:

$$\frac{1}{(s^2-a^2)^{3/2}}$$

Im not familiar with laurent series please help me

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Using $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$ we have \begin{eqnarray} \frac{1}{\sqrt{(s^2-a^2)^3}}&=&(s^2-a^2)^{-\frac32}\\ &=&s^{-3}\left(1-\left(\dfrac{a}{s}\right)^2\right)^{-\frac32}\\ &=&\frac{1}{s^3}\left(1 -\left(-\dfrac{3}{2}\right)\left(\dfrac{a}{s}\right)^2 +\frac{(-\frac{3}{2})(-\frac{3}{2}-1)}{2!}\left(\dfrac{a}{s}\right)^4 -\frac{(-\frac{3}{2})(-\frac{3}{2}-1)(-\frac{3}{2}-2)}{3!}\left(\dfrac{a}{s}\right)^6 +\cdots\right)\\ &=&1 +\dfrac{3a^2}{2s^5} +\dfrac{15a^4}{8s^7} +\dfrac{35a^6}{16s^9} +\cdots \end{eqnarray}

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    Can i ask sir what do you call the formula you used? – 2017-02-22
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    It is binomial expansion for real $n$. – 2017-02-22
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    Sir if i dont know the laplace transform of the series above how can i used binomial theorem to arrived at its laplace transform? – 2017-02-22
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    Have you solved your problem with wolfram alpha.? – 2017-02-22
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    No by using general formula, and im stuck at the series expansion because i dont know the laurent series i have just check with wolfram alpha if my answer was correct – 2017-02-22
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    Usually problems are designed to solve simply, I think you may find solution without using binomial theorem. Say your problem if you want. – 2017-02-22