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I tried solving this question but failed.

a) Expand $(1+2x)^{1/4}$ in ascending powers of $x$ up to and including the term in $x^3$, simplifying each term as far as possible.

b) By substituting $x=\frac{1}{16}$ into your expansion, obtain an approximation, to $6$ significant figures, for $8^{1/4}$.

I successfully solved question a but I failed to solve question b as $1+2(\frac{1}{16}) \ne 8$.

Is there anything wrong with the question or is it my problem?

Answer I got for question a is $$ 1 + \frac{1}{2}x - \frac{3}{2}x^2 + \frac{63}{4}x^3 + \cdots $$

Is my answer correct?

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  • $\begingroup$ Your answer is not correct. I am pretty sure the problem is that you are multiplying by $n!$ instead of dividing. It should be $1+\frac{1}{2}x-\frac{3}{8}x^2+\frac{7}{16}x^3$. $\endgroup$ – André Nicolas May 9 '13 at 6:23
  • $\begingroup$ Your coefficients $a_0$ and $a_1$ are correct, but $a_2$ and $a_3$ are incorrect. I agree that you cannot estimate $8^{1/4}$ by evaluating your series at $x = \frac{1}{16}$. $\endgroup$ – Sammy Black May 9 '13 at 6:26
  • $\begingroup$ On the other hand, if you evaluate at $-1/4$ and multiply the end result by $2$, you can get there. $\endgroup$ – Raskolnikov May 9 '13 at 6:27
  • $\begingroup$ If all you need is verification of correctness, Wolfram Alpha (link computes series $(1+2x)^{1/4})$ is a good resource. $\endgroup$ – Aryabhata May 9 '13 at 6:37

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