Calculate-the-sum-k-1-n-1-k-k-2-1-

Question Number 121950 by liberty last updated on 12/Nov/20

Calculate the sum \underset{k = 1}{\sum^{n}} \frac{1}{\sqrt{k + \sqrt{k^{2} + 1}}} .

Answered by bobhans last updated on 12/Nov/20

l e t a = k + \sqrt{k^{2} - 1} a n d b = k - \sqrt{k^{2} + 1}

t h e n w e h a v e \to {\begin{cases} a + b = 2 k \\ a b = k^{2} - {(\sqrt{k^{2} - 1})}^{2} = 1 \end{cases}

s o w e g e t \frac{1}{\sqrt{k + \sqrt{k^{2} + 1}}} = \frac{1}{\sqrt{a}} = b .

O n t h e o t h e r h a n d, o b s e r v e t h a t

b = k - \sqrt{k^{2} - 1} = k - \sqrt{(k + 1) (k - 1)}

= \frac{k + 1 + k - 1}{2} - \sqrt{(k + 1) (k - 1)}

= \frac{k + 1 - 2 \sqrt{(k + 1) (k - 1)} + k - 1}{2}

= \frac{{(\sqrt{k + 1} - \sqrt{k - 1})}^{2}}{2}

h e n c e \sqrt{b} = \frac{\sqrt{k + 1} - \sqrt{k - 1}}{\sqrt{2}}

\sqrt{b} = \frac{\sqrt{k + 1} - \sqrt{k} + \sqrt{k} - \sqrt{k - 1}}{\sqrt{2}}

n o w t h e e q u a t i o n c a n b e r e w r i t t e a s

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

b y t e l e s c o p i n g s e r i e s g i v e s

= \frac{\sqrt{n + 1} - 1}{\sqrt{2}} + \frac{\sqrt{n} - \sqrt{0}}{\sqrt{2}}

= \frac{\sqrt{n + 1} + \sqrt{n} - 1}{\sqrt{2}} .

Commented by liberty last updated on 12/Nov/20

correct and great \dots

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