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How to transform $$\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m$$ into $$\left({b\over a}\right)^n -1\over \left({b\over a}\right)^m -1$$?

Thanks.

Added: As per comments, there is an additional assumption that $a,b\gt 0$, $b=1-a$, and $b\gt a$.

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    @smanoos: I am not quite sure what "divide each expression by one of the expressions" means...?2011-10-13
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    What makes you think you can? If $b=0$, then the first expression simplifies to $a^{m-n}$ and the second expression simplifies to $1$.2011-10-13
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    @ThomasAndrews: Because I am so told... What if I add the constraint that $b>a$ and $b=1-a$ and $a,b>0$?2011-10-13
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    If $b=1-a$, then $1-4ab = 1-4(1-a)a = 1-4a+4a^2 = (1-2a)^2$; since $1-a\gt a$, then $2a\lt 1$, so $1\pm\sqrt{1-4ab} = 1\pm|1-2a| = 1 \pm (1-2a)$. So you have $$\frac{\left(\frac{2-2a}{2a}\right)^n - 1}{\left(\frac{2-2a}{2a}\right)^m - 1}$$and since $2-2a=2(1-a)=2b$, the simplification follows immediately.2011-10-13
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    @ArturoMagidin: Thanks sooo much! :D2011-10-13
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    Yeah, that's a lot of conditions to leave out of the initial problem.2011-10-13
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    @ThomasAndrews: Sorry :S2011-10-13

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