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12 May 2020, 09:48
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A
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Difficulty:
85%
(hard)
Question Stats:
53% (02:34) correct
47%
(02:27)
wrong
based on 141
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In an sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., the difference between successive terms is constant. What is the value of \(\frac{1}{\sqrt{a_1 }+ \sqrt{a_2}} + \frac{1}{\sqrt{a_2 }+ \sqrt{a_3}} + ...+ \frac{1}{\sqrt{a_n }+ \sqrt{a_{n+1}}} \)?
A. \(\frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}}\)
B. \(\frac{n-1}{\sqrt{a_1} + \sqrt{a_{n}}}\)
C. \(\frac{n}{\sqrt{a_1} - \sqrt{a_{n+1}}}\)
D. \(\frac{n-1}{\sqrt{a_1} - \sqrt{a_{n+1}}}\)
E. \(\frac{n-1}{\sqrt{a_1} - \sqrt{a_{n}}}\)
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}}\)
B. \(\frac{n-1}{\sqrt{a_1} + \sqrt{a_{n}}}\)
C. \(\frac{n}{\sqrt{a_1} - \sqrt{a_{n+1}}}\)
D. \(\frac{n-1}{\sqrt{a_1} - \sqrt{a_{n+1}}}\)
E. \(\frac{n-1}{\sqrt{a_1} - \sqrt{a_{n}}}\)
Are You Up For the Challenge: 700 Level Questions
12 May 2020, 10:20
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Bunuel
Given: In an sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., the difference between successive terms is constant.
Asked: What is the value of \(\frac{1}{\sqrt{a_1 }+ \sqrt{a_2}} + \frac{1}{\sqrt{a_2 }+ \sqrt{a_3}} + ...+ \frac{1}{\sqrt{a_n }+ \sqrt{a_{n+1}}} \)?
\(\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}} =\frac{ \sqrt{a_n} - \sqrt{a_{n-1}} }{ d} \)
\(d = \frac{(a_{n+1} - a_1) }{n}\)
\(\frac{1}{\sqrt{a_1 }+ \sqrt{a_2}} + \frac{1}{\sqrt{a_2 }+ \sqrt{a_3}} + ...+ \frac{1}{\sqrt{a_n }+ \sqrt{a_{n+1}}} = \)
\(=\frac{1}{d} * (\sqrt{a_2} - \sqrt{a_1} + \sqrt{a_3} - \sqrt{a_2} + .... + \sqrt{a_n} - \sqrt{a_{n-1}} +\sqrt{a_{n+1}} - \sqrt{a_n}) = \)
\(=\frac{1}{d}*(\sqrt{a_{n+1}} - \sqrt{a_1}) = \frac{n}{(a_{n+1} - a_1)} *(\sqrt{a_{n+1}} - \sqrt{a_1}) =\)
\(= \frac{n}{(\sqrt{a_{n+1}} + \sqrt{a_1})} \)
IMO A
