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I need to find the values of $a$ and $n$ in the following

$$ (1 + ax)^n = \sum_{i=0}^\infty \binom{2i}{i}x^i$$

How can I compare things and show that $a=-4$ and $n = -{1 \over 2}$. It's easy to expand $1 \over \sqrt{1 -4x}$ see that it's valid but I'm looking other way around.

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    $\begingroup$ You can see that $n$ is not an integer. If it were, then the left hand side would be a polynomial, while the right hand side is an infinite sum of monomials. $\endgroup$ – Michael Albanese Oct 2 '12 at 9:30
  • $\begingroup$ yeah i know that ... but how to show that $n=-1/2$ $\endgroup$ – Santosh Linkha Oct 2 '12 at 9:45
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    $\begingroup$ @Michael: You can see $n$ is not a positive integer $\endgroup$ – Henry Oct 2 '12 at 10:05
  • $\begingroup$ Looking for another way around is rather strange. Unless you've already studied the series in the RHS quite a bit, there is no reason to assume it would be a (fractional) power of a binomial at all, or even satisfy any algebraic equation involving polynomials. $\endgroup$ – Marc van Leeuwen Oct 2 '12 at 10:17
  • $\begingroup$ @Henry: Yes, you are correct, I missed that. $\endgroup$ – Michael Albanese Oct 2 '12 at 10:18
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$$(1+ax)^{n}=\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}=\sum_{n=0}^{\infty}\frac{(2n)!}{n!n!}x^{n}$$ Derivated at both sides, we get the first equation: $$na(1+ax)^{n-1}=\sum_{n=1}^{\infty}\frac{n(2n)!}{n!n!}x^{n-1}$$ Derivated at both sides of the first equation, we get the second equation: $$n(n-1)a^{2}(1+ax)^{n-2}=\sum_{n=2}^{\infty}\frac{n(n-1)(2n)!}{n!n!}x^{n-2}$$ Let $x=0$ in above two equations, we can get that: $$na=2\,\,\text{and}\,\,n(n-1)a^{2}=12$$ then solve the equation and we can get that $a=-4$ and $n=-\frac{1}{2}$.

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  • $\begingroup$ It would be a good thing to note that, while we are using the same variable names (namely $n$) both on the left and on the right side of the equations, for one to derive that $na = 2$ and $n(n-1)a^2 = 12$, as you did, you have to consider that, on the RHS, when setting $x = 0$, you need the coefficients to be already known to be constant---not involving the variable $n$ of the LHS. $\endgroup$ – rbrito Nov 14 '12 at 14:56
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At first, consider the series $\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}$, as $\frac{C_{2n+2}^{n+1}}{C_{2n}^{n}}=\frac{4(2n+1)}{2n+2}$, we can judge that the series convergent in $[-\frac{1}{4},\frac{1}{4})$.

Note $f(x)=\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}=\sum_{n=0}^{\infty}\frac{(2n)!}{n!n!}x^{n}$, then $f(0)=1$, $$f^{\prime}(x) =\sum_{n=1}^{\infty}\frac{(2n)!}{n!(n-1)!}x^{n-1} =\sum_{n=0}^{\infty}\frac{(2n+2)!}{(n+1)!n!}x^{n} =\sum_{n=0}^{\infty}\frac{(4n+2)(2n)!}{n!n!}x^{n}$$ $$xf^{\prime}(x) =\sum_{n=1}^{\infty}\frac{(2n)!}{n!(n-1)!}x^{n} =\sum_{n=0}^{\infty}\frac{n(2n)!}{n!n!}x^{n}$$ $$(1-4x)f^{\prime}(x)=2\sum_{n=0}^{\infty}\frac{(2n)!}{n!n!}x^{n}=2f(x)$$ then solve the last ordinary differential equation, we get $f(x)=\frac{c}{\sqrt{1-4x}}$, as $f(0)=1$, so $f(x)=\frac{1}{\sqrt{1-4x}}$, i.e. $\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}=\frac{1}{\sqrt{1-4x}}, x\in[-\frac{1}{4},\frac{1}{4})$.

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You can show that the sum is the solution of a differerential equation (try to express the derivative in function of the sum itself).

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    $\begingroup$ could you please elaborate!! $\endgroup$ – Santosh Linkha Oct 2 '12 at 9:09

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