An Application of Quadratic Formula

Here's a curious fact:

\(\displaystyle \bigg(\frac{\sqrt{2+\sqrt{1}}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{1}}+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}}-1\bigg)^2=2. \)

It would be a risky venture to prove the unappealing identity directly. Interestingly, it is only a special case of a more general equality that motivated the authors of a problem from the 2009 Japanese Mathematical Olympiad [Lecture Notes, p 30]:

Evaluate

\(\displaystyle \frac{\displaystyle\sum_{n=1}^{99}(10+\sqrt{n})}{\displaystyle\sum_{n=1}^{99}(10-\sqrt{n})} \)

As a matter of fact, the following identity holds:

For \(N\gt 1\),

\(\displaystyle \frac{\displaystyle\sum_{n=1}^{N^{2}-1}(N+\sqrt{n})}{\displaystyle\sum_{n=1}^{N^{2}-1}(N-\sqrt{n})} = 1+\sqrt{2}. \)

The identity at the top is obtained by letting \(N=2\). I have a feeling that the more general formulation may be more suggestive of a possible solution. The choice of a specific value for \(N\), be it \(10\) or \(2\) seems (to me at least) to make the problem more obscure. Of course it is wonderful that the expression is independent of \(N\).

Solution

References

  1. Xu Jiagu, Lecture Notes on Mathematical Olympiad Courses, v 8, (For senior section, v 2), World Scientific, 2012

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Copyright © 1996-2017 Alexander Bogomolny

For \(N\gt 1\),

\(\displaystyle \frac{\displaystyle\sum_{n=1}^{N^{2}-1}(N+\sqrt{n})}{\displaystyle\sum_{n=1}^{N^{2}-1}(N-\sqrt{n})} = 1+\sqrt{2}. \)

The starting point for a solution is the following observation:, for positive \(a\) and \(b\),

\(\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}.\)

In that formula, \(a\) and \(b\) can be sought of as the roots of the quadratic equation

\(x^{2}-(a+b)x+ab = 0.\)

Such a view will prove helpful, provided we can find \(a\) and \(b\) that satisfy Viète's formulas, \(a+b=2N\) and \(ab=n\). So we check the quadratic equation

\(x^{2}-2Nx+n = 0.\)

The discriminant of the equation equals \(4(N^{2}-n)\) and is positive exactly when \(1\le n \lt N^2\). The two roots are \(N\pm\sqrt{N^{2}-n}\). Taking \(a=N+\sqrt{N^{2}-n}\) and \(b=N-\sqrt{N^{2}-n}\) we see that,

\( \sqrt{N+\sqrt{N^{2}-n}}+\sqrt{N-\sqrt{N^{2}-n}} = \sqrt{2N+2\sqrt{n}}. \)

Thus we define, say, \(\displaystyle S=\sum_{n=1}^{N^{2}-1}(N+\sqrt{n})\) and \(\displaystyle T=\sum_{n=1}^{N^{2}-1}(N-\sqrt{n})\). What we just found shows that

\(\displaystyle \begin{align} \sqrt{2}S &= \sum_{n=1}^{N^{2}-1}(\sqrt{N+\sqrt{N^{2}-n}}+\sqrt{N-\sqrt{N^{2}-n}}) \\ &= \sum_{n=N^{2}-1}^{1}(\sqrt{N+\sqrt{N^{2}-n}}+\sqrt{N-\sqrt{N^{2}-n}}) \\ &= \sum_{n=1}^{N^{2}-1}(\sqrt{N+\sqrt{n}}+\sqrt{N-\sqrt{n}}) \\ &= S + T, \end{align} \)

which implies that \(\displaystyle\frac{S}{T}=\frac{1}{\sqrt{2}-1}=1+\sqrt{2}\).

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Copyright © 1996-2017 Alexander Bogomolny

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