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Consider $$f(x) = \sum^{\infty}_{n=0} \frac{4n-7}{6n+7} x^n.$$

Find $ f'(x).$ I simply took the derivative which I thought is $$\sum^{\infty}_{n=0} \frac{4n^2-7n}{6n+7} x^{n-1}.$$

The response says: "This is a very subtle mistake. You have included the $n=0$ term but you should have removed it and then reindexed.

The right answer they claim is: $$ \sum^{\infty}_{n=0} \frac{4n^2+n-3}{6n+13} x^n.$$

I don't understand what I did wrong and what they want me to do?

Ѕᴀᴀᴅ
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Haim
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  • You can reindex, but the answer given is not correct in the denominator. – Paul Apr 27 '17 at 01:18
  • I would suggest just trying to write out a few terms to see why they're equivalent (they're both exactly the same, but yours includes an unnecessary $0$ term when $n=0$) – user12345 Apr 27 '17 at 01:50

1 Answers1

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Note simply that

$$\sum_{n=0}^\infty \frac{n(4n-7)}{6n+7}\,x^{n-1}=\sum_{n=1}^\infty \frac{n(4n-7)}{6n+7}\,x^{n-1}\tag 1$$

since the first term (i.e., $n=0$) of the series on the left-hand side of $(1)$ is $0$.

Now enforcing the substitution $n\to n+1$ in the series on the right-hand side of $(1)$ yields

$$\sum_{n=0}^\infty \frac{n(4-7)}{6n+7}\,x^{n-1}=\sum_{n=0}^\infty \frac{(n+1)(4(n+1)-7)}{6(n+1)+7}\,x^{n}=\sum_{n=0}^\infty \frac{4n^2+n-3}{6n+13}\,x^{n}$$

Mark Viola
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