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Question Number 149342 by MJS_new last updated on 04/Aug/21

is this true and is there a proof ?

\forall k \in N ∣ k > 1

\frac{\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k + \sqrt{n}}}{\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k - \sqrt{n}}} = 1 + \sqrt{2}

Commented by mr W last updated on 04/Aug/21

y e s, {i t}^{'} s t r u e .

Answered by mr W last updated on 04/Aug/21

p r o o f :

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

\sqrt{k + \sqrt{n}} - \sqrt{k - \sqrt{n}} = \sqrt{2} \times \sqrt{k - \sqrt{k^{2} - n}}

\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k + \sqrt{n}} - \underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k - \sqrt{n}} = \sqrt{2} \times \underset{n = k^{2} - 1}{\sum^{1}} \sqrt{k - \sqrt{k^{2} - n}}

\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k + \sqrt{n}} - \underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k - \sqrt{n}} = \sqrt{2} \times \underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k - \sqrt{n}}

\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k + \sqrt{n}} = (\sqrt{2} + 1) \underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k - \sqrt{n}}

\Rightarrow \frac{\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k + \sqrt{n}}}{\underset{n = 1}{\sum^{k^{2} - 1}} \sqrt{k - \sqrt{n}}} = \sqrt{2} + 1

Commented by MJS_new last updated on 04/Aug/21

thank you!

Commented by Tawa11 last updated on 06/Nov/21

great

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