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ข้อความเดิมเขียนโดยคุณ ~ToucHUp~
3.จงหาค่าของ $ \sum_{n = 1}^{360}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n} } $
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ลองแทนค่า $n = 1, 2, 3, 4 \ $ เพื่อดูรูปแบบ
$\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n} } + \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n} } + \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n} } + \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n} } $
$ = \frac{1}{1\sqrt{1+1}+(1+1)\sqrt{1} } + \frac{1}{2\sqrt{2+1}+(2+1)\sqrt{2} } + \frac{1}{3\sqrt{3+1}+(3+1)\sqrt{3} } + \frac{1}{4\sqrt{4+1}+(4+1)\sqrt{4} } $
$ = \frac{1}{\sqrt{2}+2 } + \frac{1}{2\sqrt{3}+3\sqrt{2} } + \frac{1}{3\sqrt{4} +4\sqrt{3} } + \frac{1}{4\sqrt{5}+5\sqrt{4} }$
$ = \frac{\sqrt{2} -2}{2-4} + \frac{2\sqrt{3}-3\sqrt{2} }{12-18} + \frac{3\sqrt{4}-4\sqrt{3} }{36-48} + \frac{4\sqrt{5} -5\sqrt{4} }{80-100}$
$ = \frac{2-\sqrt{2} }{2} + \frac{3\sqrt{2}-2\sqrt{3} }{6} +\frac{4\sqrt{3}-3\sqrt{4} }{12} + \frac{5\sqrt{4}-4\sqrt{5} }{20}$
$ = \frac{2}{2} - \frac{\sqrt{2} }{2} + \frac{3\sqrt{2} }{6} - \frac{2\sqrt{3} }{6} +\frac{4\sqrt{3} }{12} - \frac{3\sqrt{4} }{12} + \frac{5\sqrt{4} }{20} -\frac{4\sqrt{5} }{20}$
$ =1 - \frac{\sqrt{2} }{2} + \frac{\sqrt{2} }{2} - \frac{\sqrt{3} }{3} + \frac{\sqrt{3} }{3} - \frac{\sqrt{4} }{4} + \frac{\sqrt{4} }{4} - \frac{\sqrt{5} }{5}$
$ = 1 - \frac{\sqrt{5} }{5}$
$\therefore \ \ \sum_{n = 1}^{360}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n} } = 1 - \frac{\sqrt{361} }{361} = 1 - \frac{19}{361} = \frac{18}{19}$