Given-a-n-1-1-1-n-2-1-1-1-n-2-The-value-of-n-1-2015-4-a-n-is-

Question Number 115897 by Joel574 last updated on 29/Sep/20

Given

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

The value of \underset{n = 1}{\sum^{2015}} (\frac{4}{a_{n}}) is \dots

Commented by Joel574 last updated on 29/Sep/20

(a) \sqrt{2015^{2} + 2016^{2}}

(b) \sqrt{2015^{2} + 2016^{2}} + 1

(c) \sqrt{2014^{2} + 2015^{2}} + 1

(d) \sqrt{2015^{2} + 2016^{2}} - 1

(e) \sqrt{2014^{2} + 2015^{2}} - 1

Answered by mindispower last updated on 29/Sep/20

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

\frac{1}{a_{n}} = \frac{n}{\sqrt{2 n^{2} + 2 n + 1} + \sqrt{2 n^{2} - 2 n + 1}}

= \frac{n}{4 n} (\sqrt{2 n^{2} + 2 n + 1} - \sqrt{2 n^{2} - 2 n + 1})

= \frac{1}{4} (\sqrt{2 n^{2} + 2 n + 1} - \sqrt{2 n^{2} - 2 n + 1})

u_{n} = \sqrt{2 n^{2} - 2 n + 1}

u_{n + 1} = \sqrt{2 {(n + 1)}^{2} - 2 (n + 1) + 1}

= \sqrt{2 n^{2} + 2 n + 1}

\frac{1}{a n} = \frac{1}{4} (U_{n + 1} - U_{n})

Σ \frac{1}{a_{n}} = \frac{1}{4} Σ (u_{n + 1} - u_{n})

\sum_{n ⩽ 2015} \frac{4}{a_{n}} = U_{2016} - u_{1}

u_{1} = 1

u_{n} = \sqrt{{(n + 1)}^{2} + n^{2}}

u_{2016} = \sqrt{2017^{2} + 2016^{2}}

\Rightarrow Σ \frac{4}{a_{n}} = \sqrt{2017^{2} + 2016^{2}} - 1

Commented by Joel574 last updated on 30/Sep/20

T h a n k y o u v e r y m u c h

Answered by Dwaipayan Shikari last updated on 29/Sep/20

\underset{n = 1}{\sum^{n}} (\frac{4}{a_{n}}) = \underset{n = 1}{\sum^{n}} \frac{4}{\sqrt{1 + {(1 - \frac{1}{n})}^{2}} + \sqrt{1 + {(1 + \frac{1}{n})}^{2}}}

\underset{n = 1}{\sum^{n}} 4 \frac{\sqrt{1 + {(1 - \frac{1}{n})}^{2}} - \sqrt{1 + {(1 + \frac{1}{n})}^{2}}}{- \frac{4}{n}}

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

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

= \sqrt{5} - \sqrt{1} + \sqrt{13} - \sqrt{5} + \dots + \sqrt{n^{2} + {(n + 1)}^{2}} - \sqrt{{(n - 1)}^{2} + n^{2}}

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

Commented by Joel574 last updated on 30/Sep/20

T h a n k y o u v e r y m u c h

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