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I need to simplify $\frac{2n+1}{n^2(n+1^2)}$ as part of an exam question. The solution states $$\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$$ In the solution it does not state how this simplification was made, I figured this could be done in quite a long winded fashion, using partial fractions. But from how it's written in the solutions it seems like this should be an easy simplification.

Is there a simple trick to simplifying fractions like this?

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  • $\begingroup$ This likely uses partial fractions, yes $\endgroup$
    – 123
    Commented Jan 13, 2015 at 18:09
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    $\begingroup$ There is a simple simplification, but usually one finds it having in mind the final form after partial fractions, even if one doesn't even think about partial fractions. Having the answer, to find the simplification, simply start from the RHS and prove it equals the LHS, the trick will present itself. $\endgroup$
    – Git Gud
    Commented Jan 13, 2015 at 18:12

3 Answers 3

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HINT

Use Partial Fraction Decomposition,

$$\frac{2n+1}{n^2(n+1)^2}=\frac An+\frac B{n^2}+\frac C{n+1}+\frac D{(n+1)^2}$$

$$\iff2n+1=An(n+1)^2+B(n+1)^2+Cn^2(n+1)+Dn^2$$

Now compare the constants & the coefficients of $n,n^2,n^3$ to find $A,B,C,D$


Alternatively by observation,

$$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{(n+1)^2n^2}=\cdots$$

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  • $\begingroup$ That's not much of a hint, given that the question was exactly whether there's an easier way than this. $\endgroup$ Commented Jan 13, 2015 at 18:18
  • $\begingroup$ @HenningMakholm, I'm eager to see an easier method than my alternative approach $\endgroup$ Commented Jan 13, 2015 at 18:19
  • $\begingroup$ But the alternative isn't even a method. It's just a proof that the conclusion is right -- not a forward way of discovering that simplification. $\endgroup$ Commented Jan 13, 2015 at 18:21
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Note that $$\frac{2n+1}{n^2(n+1)^2} = \frac{(n^2 + 2n + 1)-n^2}{n^2(n+1)^2}$$ $$=\frac{(n+1)^2-n^2}{n^2(n+1)^2}$$ $$=\frac{(n+1)^2}{n^2(n+1)^2}-\frac{n^2}{n^2(n+1)^2}$$ $$=\frac1{n^2}-\frac1{(n+1)^2}$$ as desired. This approach is motivated by noticing that the original numerator is "almost" a recognizable perfect square, so one is tempted to add and subtract the missing quantity that completes the square.

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  • $\begingroup$ Have you noticed my alternative method? $\endgroup$ Commented Jan 13, 2015 at 18:19
  • $\begingroup$ @labbhattacharjee: Didn't see it until after I posted. And now I see that yet another copy has been posted by amWhy. $\endgroup$
    – MPW
    Commented Jan 13, 2015 at 18:22
  • $\begingroup$ MPW: Sorry about that; I was composing, I suspect, while you posted. $\endgroup$
    – amWhy
    Commented Jan 13, 2015 at 18:26
  • $\begingroup$ The final paragraph is exactly what I was hoping for. Thank you! $\endgroup$
    – Elis Jones
    Commented Jan 13, 2015 at 18:27
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use that $\frac{2n+1}{n^2(n+1)^2}=\frac{A}{n}+\frac{B}{n^2}+\frac{C}{n+1}+\frac{D}{(n+1)^2}$

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    $\begingroup$ It seems that the OP is asking if there is some tricky method to avoid partial fractions. $\endgroup$
    – robjohn
    Commented Jan 13, 2015 at 18:15

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