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# A function $f$ is defined by $f(x) = 1 + 2x + x^2 + 2x^3 + x^4 + \cdot \cdot \cdot$that is, its coefficients are $c_{2n} = 1$ and $c_{2n + 1} = 2$ for all $n \ge 0.$ Find the interval of convergence of the series and find an explicit formula for $f(x).$

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

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okay, if we rewrite this, this is one plus X squared plus X to the fourth. Plus that that that And then we also have these two x to execute etcetera terms. We can factor out a two x from all those guys and write it as two acts times one plus x squared plus X to the fourth. Plus that that. Okay, so remember that one plus x plus X squared plus dot, dot dot is one over one minus x. Here. We're doing the same thing, except instead of using acts were using X squared. So this is going to be one over one minus x squared, and then, similarly, over here, this is too X divided by one minus X squared. So we have two x plus one over one minus x squared. Okay, So this is the explicit formula that we have and the interval of convergence for this whole sum is going to be the same thing as the interval convergence for this guy here, right. We can rewrite this whole thing as two x plus one quantity multiplied by this infinite sum. So this infinite sum is the only thing that we have to really worry about here. Right? And this infinite sum is just X to the two n from any Kools zero to infinity. Okay, so you can use the ratio test. Figure out what this is in case you forgot what this radius of convergence is here. Using the ratio test, we would get absolute value of X to the two n plus one divided by X to the two end. So this is going to turn out to just be X squared. And to get that to be less than one acts would be between minus one and one. The radius of convergence here is one and our values of accer between minus one and one. So now we just need to check the end points. So we plug in X equals minus one. Here we get one plus one plus one plus one. Is that all? Certainly be divergent if you plug in one here. Same thing. One plus one plus one plus one divergent. So our interval of convergence is minus one. The one not including them. So open in trouble. Here

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