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I am required to find the co-efficients of this power series: $2x\ln(1+2x)$

I approached the problem by considering the $\ln(1+2x)$ part as the integral of $2/(1+2x)$ and applied the geometric series to this. Factoring the $2$ on the numerator outside this expression gives us $4x$ on the outside (after multiplying it with $2x$).

I need to get the first 5 nonzero co-efficients, and what I got is: $0, 4, -8, 16$ and $-32$. Only $0$ is right, according to the online portal.

How do I approach this problem?

MonkeyKing
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ShinyPebble
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    You could also calculate the 2x and ln part separately and than multiply – Quality Apr 12 '15 at 05:15
  • You didn't integrate. Your mistake would have been clearer if you had written out what you tried in more detail. – user21820 Apr 12 '15 at 05:18
  • You should have $ \ \ln (1+2x) \ = \sum_{n=1}^{\infty} \ (-1)^{n-1} \ \frac{(2x)^n}{n} \ $ as an intermediate result. I think you may also have an excess "2" floating around in there... – colormegone Apr 12 '15 at 05:22

2 Answers2

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$\log(1+2x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2x)^n}{n}$ implies that

$$2x\log(1+2x)=\sum_{n=1}^{\infty} \frac{(-2x)^{n+1} }{n}=\sum_{n=2}^{\infty} \frac{(-2x)^{n} }{n-1}$$

Mark Viola
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I am not sure what you are talking about in the second paragraph. However since you are doing it with geometric series, then you know that $\frac{1}{1+2x} = \sum^{\infty}_{n=1}(-2x)^{n-1}$. Then you just integrate and multiply by $4x$, which gives $2x\ln(1+2x) = 4x \sum^{\infty}_{n=1} \frac{(-2)^{n-1}x^{n}}{n} = \sum^{\infty}_{n=1}\frac{(-2x)^{n+1}}{n}$.

MonkeyKing
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  • Thanks, that did help. I had made an error in my calculations, but seeing the summation form helped me do it again correctly. – ShinyPebble Apr 12 '15 at 06:01