Find-the-sum-1-1-2-2-1-3-2-1-1-3-2-1-4-2-1-1-999-2-1-1000-2-

Question Number 91068 by Maclaurin Stickker last updated on 27/Apr/20

F i n d t h e s u m

\sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + \sqrt{1 + \frac{1}{3^{2}} + \frac{1}{4^{2}}} + \dots + \sqrt{1 + \frac{1}{999^{2}} + \frac{1}{1000^{2}}}

Answered by mr W last updated on 28/Apr/20

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

\underset{n = 2}{\sum^{999}} \sqrt{1 + \frac{1}{n^{2}} + \frac{1}{{(n + 1)}^{2}}}

= 998 + \frac{1}{2} - \frac{1}{1000}

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