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23 Nov 2021, 08:42
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If \(\frac{1}{5\sqrt{4} + 4\sqrt{5} } + \frac{1}{6\sqrt{5} + 5\sqrt{6} } +\frac{1}{7\sqrt{6} + 6\sqrt{7} } +...+\frac{1}{n\sqrt{n-1} + (n-1)\sqrt{n} }=\frac{3}{10}\), then what is the value of \(n\)?
A. \(9\)
B. \(10\)
C. \(15\)
D. \(20\)
E. \(25\)
A. \(9\)
B. \(10\)
C. \(15\)
D. \(20\)
E. \(25\)
23 Nov 2021, 08:42
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Official Solution:
If \(\frac{1}{5\sqrt{4} + 4\sqrt{5} } + \frac{1}{6\sqrt{5} + 5\sqrt{6} } +\frac{1}{7\sqrt{6} + 6\sqrt{7} } +...+\frac{1}{n\sqrt{n-1} + (n-1)\sqrt{n} }=\frac{3}{10}\), then what is the value of \(n\)?
A. \(9\)
B. \(10\)
C. \(15\)
D. \(20\)
E. \(25\)
Rationalize by multiplying the denominator and the numerator of each fraction by \(\frac{1}{x\sqrt{x-1} - (x-1)\sqrt{x} }\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator):
\(\frac{5\sqrt{4} - 4\sqrt{5} }{(5\sqrt{4} + 4\sqrt{5})(5\sqrt{4} - 4\sqrt{5})} + \frac{6\sqrt{5} - 5\sqrt{6} }{(6\sqrt{5} + 5\sqrt{6})(6\sqrt{5} - 5\sqrt{6})} +\frac{7\sqrt{6} - 6\sqrt{7} }{(7\sqrt{6} + 6\sqrt{7})(7\sqrt{6} - 6\sqrt{7})} +...+\frac{n\sqrt{n-1} - (n-1)\sqrt{n} }{(n\sqrt{n-1} + (n-1)\sqrt{n})(n\sqrt{n-1} - (n-1)\sqrt{n})}=\frac{3}{10}\);
\(\frac{5\sqrt{4} - 4\sqrt{5} }{20} + \frac{6\sqrt{5} - 5\sqrt{6} }{30} +\frac{7\sqrt{6} - 6\sqrt{7} }{42} +...+\frac{n\sqrt{n-1} - (n-1)\sqrt{n} }{n(n-1)}=\frac{3}{10}\)
\((\frac{5\sqrt{4} }{20} - \frac{4\sqrt{5} }{20}) + (\frac{6\sqrt{5} }{30} - \frac{5\sqrt{6} }{30}) +(\frac{7\sqrt{6} }{42} - \frac{6\sqrt{7} }{42}) +...+(\frac{n\sqrt{n-1} }{n(n-1)} - \frac{(n-1)\sqrt{n} }{n(n-1)})=\frac{3}{10}\);
\((\frac{\sqrt{4} }{4} - \frac{\sqrt{5} }{5}) + (\frac{\sqrt{5} }{5} - \frac{\sqrt{6} }{6}) +(\frac{\sqrt{6} }{6} - \frac{\sqrt{7} }{7}) +...+(\frac{\sqrt{n-1} }{(n-1)} - \frac{\sqrt{n} }{n})=\frac{3}{10}\);
Notice that everything except the first and the last terms will cancel each other, and we'll be left with \(\frac{\sqrt{4} }{4}- \frac{\sqrt{n} }{n}=\frac{3}{10}\)
\(\frac{1}{2} - \frac{1}{\sqrt{n} } = \frac{3}{10}\);
\(\sqrt{n}=5\);
\(n=25\).
Answer: E
If \(\frac{1}{5\sqrt{4} + 4\sqrt{5} } + \frac{1}{6\sqrt{5} + 5\sqrt{6} } +\frac{1}{7\sqrt{6} + 6\sqrt{7} } +...+\frac{1}{n\sqrt{n-1} + (n-1)\sqrt{n} }=\frac{3}{10}\), then what is the value of \(n\)?
A. \(9\)
B. \(10\)
C. \(15\)
D. \(20\)
E. \(25\)
Rationalize by multiplying the denominator and the numerator of each fraction by \(\frac{1}{x\sqrt{x-1} - (x-1)\sqrt{x} }\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator):
\(\frac{5\sqrt{4} - 4\sqrt{5} }{(5\sqrt{4} + 4\sqrt{5})(5\sqrt{4} - 4\sqrt{5})} + \frac{6\sqrt{5} - 5\sqrt{6} }{(6\sqrt{5} + 5\sqrt{6})(6\sqrt{5} - 5\sqrt{6})} +\frac{7\sqrt{6} - 6\sqrt{7} }{(7\sqrt{6} + 6\sqrt{7})(7\sqrt{6} - 6\sqrt{7})} +...+\frac{n\sqrt{n-1} - (n-1)\sqrt{n} }{(n\sqrt{n-1} + (n-1)\sqrt{n})(n\sqrt{n-1} - (n-1)\sqrt{n})}=\frac{3}{10}\);
\(\frac{5\sqrt{4} - 4\sqrt{5} }{20} + \frac{6\sqrt{5} - 5\sqrt{6} }{30} +\frac{7\sqrt{6} - 6\sqrt{7} }{42} +...+\frac{n\sqrt{n-1} - (n-1)\sqrt{n} }{n(n-1)}=\frac{3}{10}\)
\((\frac{5\sqrt{4} }{20} - \frac{4\sqrt{5} }{20}) + (\frac{6\sqrt{5} }{30} - \frac{5\sqrt{6} }{30}) +(\frac{7\sqrt{6} }{42} - \frac{6\sqrt{7} }{42}) +...+(\frac{n\sqrt{n-1} }{n(n-1)} - \frac{(n-1)\sqrt{n} }{n(n-1)})=\frac{3}{10}\);
\((\frac{\sqrt{4} }{4} - \frac{\sqrt{5} }{5}) + (\frac{\sqrt{5} }{5} - \frac{\sqrt{6} }{6}) +(\frac{\sqrt{6} }{6} - \frac{\sqrt{7} }{7}) +...+(\frac{\sqrt{n-1} }{(n-1)} - \frac{\sqrt{n} }{n})=\frac{3}{10}\);
Notice that everything except the first and the last terms will cancel each other, and we'll be left with \(\frac{\sqrt{4} }{4}- \frac{\sqrt{n} }{n}=\frac{3}{10}\)
\(\frac{1}{2} - \frac{1}{\sqrt{n} } = \frac{3}{10}\);
\(\sqrt{n}=5\);
\(n=25\).
Answer: E
