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17 Aug 2021, 08:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:18) correct
66%
(02:19)
wrong
based on 353
sessions
History
Date
Time
Result
Not Attempted Yet
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
M37-40
A. 9
B. 10
C. 15
D. 20
E. 25
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M37-40
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17 Aug 2021, 08:49
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is it E?
I think it should be 3/10 instead of 7/10. Sorry if I am wrong, please let me know, or maybe I made some silly calculation mistake? But I rechecked 3 times.
In such questions we take conjugate of denominator and multiply both denominator and numerator with it, such that denominator becomes an integer free of roots.
In attached screenshot, we see that we converted the terms into additive terms such that all terms get cancelled out except the first and last term.
So 1/2 - 1/rootn = 7/10. Here root n is yielded as -5. If question said 3/10, we'd get n = 25.
I am not sure what wrong I did, but I am certain I have wrong answer with 7/10 on right hand side.
Anyway, my best bet would be 25 in such case, as n will have to be a perfect square regardless and if it were 1/2 + 1/rootn = 7/10, n would be 25.
I think it should be 3/10 instead of 7/10. Sorry if I am wrong, please let me know, or maybe I made some silly calculation mistake? But I rechecked 3 times.
In such questions we take conjugate of denominator and multiply both denominator and numerator with it, such that denominator becomes an integer free of roots.
In attached screenshot, we see that we converted the terms into additive terms such that all terms get cancelled out except the first and last term.
So 1/2 - 1/rootn = 7/10. Here root n is yielded as -5. If question said 3/10, we'd get n = 25.
I am not sure what wrong I did, but I am certain I have wrong answer with 7/10 on right hand side.
Anyway, my best bet would be 25 in such case, as n will have to be a perfect square regardless and if it were 1/2 + 1/rootn = 7/10, n would be 25.
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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
