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sequence of numbers a1, a2, a3, . . . , an [#permalink]
04 Jan 2016, 06:20
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38% (02:14) wrong based on 86 sessions
The sequence of numbers \(a_1\), \(a_2\), \(a_3\), . . . , \(a_n\), . . . is defined by \(a_n\)\(= \frac{1}{n}  \frac{1}{(n+2)}\) for each integer \(n ≥ 1\). What is the sum of the first 20 terms of this sequence? A. \((1+\frac{1}{2})  \frac{1}{20}\) B. \((1+\frac{1}{2})  (\frac{1}{21}+\frac{1}{22})\) C. \(1  (\frac{1}{20}+\frac{1}{22})\) D. \(1  \frac{1}{22}\) E. \(\frac{1}{20}  \frac{1}{22}\) Practice Questions Question: 12 Page: 158 Difficulty: hard
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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
04 Jan 2016, 08:40
Here we have \(an= \frac{1}{n}  \frac{1}{(n+2)}\). The idea is to find a pattern in the sum \(a1= \frac{1}{1}  \frac{1}{3}\) \(a2= \frac{1}{2}  \frac{1}{4}\) \(a3= \frac{1}{3}  \frac{1}{5}\) \(a4= \frac{1}{4}  \frac{1}{6}\) Now we have the first term from a3 can cancels second term from previous numbers Now observing the last few terms... \(a18= \frac{1}{18}  \frac{1}{20}\) \(a19= \frac{1}{19}  \frac{1}{21}\) \(a20= \frac{1}{20}  \frac{1}{22}\) Now we have here the last two terms namely \(\frac{1}{21}\) and \(\frac{1}{22}\) are not cancelled. \((1+\frac{1}{2})  (\frac{1}{21}+\frac{1}{22})\) Hence the sum is
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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
29 Mar 2016, 09:34
Now we have the first term from a3 can cancels second term from previous numbers ^^^ what do you mean by this exactly, like what exact numbers cancel?
does the sum of the first two cancel, or something else?



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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
13 Jul 2016, 13:31
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sagnik242 wrote: Now we have the first term from a3 can cancels second term from previous numbers ^^^ what do you mean by this exactly, like what exact numbers cancel?
does the sum of the first two cancel, or something else? In the first set, you cancel both the 1/3, the 1/4, and the 1/5. You see the pattern now? This cancellation will continue until the last two terms (1/21 and 1/20 cannot be cancelled because n=20 is the limit). In the first set you are left with 1/1 and 1/2. By the second set there's only 1/21 and 1/22. You add the first set and subtract the second. (1/1 + 1/2)  (1/21 + 1/22).



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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
09 Feb 2019, 08:17
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sandy wrote: The sequence of numbers a1, a2, a3, . . . , an, . . . is defined by \(an= \frac{1}{n}  \frac{1}{(n+2)}\) for each integer n ≥ 1. What is the sum of the first 20 terms of this sequence?
A. \((1+\frac{1}{2})  \frac{1}{20}\) B. \((1+\frac{1}{2})  (\frac{1}{21}+\frac{1}{22})\) C. \(1  (\frac{1}{20}+\frac{1}{22})\) D. \(1  \frac{1}{22}\) E. \(\frac{1}{20}  \frac{1}{22}\)
Applying the formula, we get: a1 = (1/1  1/3) a2 = (1/2  1/4) a3 = ( 1/3  1/5) a4 = ( 1/4  1/6) . . . . a17 = (1/17  1/19) a18 = (1/18  1/20) a19 = ( 1/19  1/21) a20 = ( 1/20  1/22) So, the SUM = (1/1  1/3) + (1/2  1/4) + (1/3  1/5) + (1/4  1/6) . . . (1/17  1/19) + (1/18  1/20) + (1/19  1/21) + (1/20  1/22)Notice that all of the SAME COLORED fractions cancel out. For example, (1/3) + 1/3 = 0The only fractions the DON'T get canceled out are those at the beginning and end of the sequence. So, SUM = 1/1 + 1/2  1/21  1/22We can rewrite this as: SUM = (1/1 + 1/2)  (1/21 + 1/22)Answer: B Cheers, Brent
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Re: sequence of numbers a1, a2, a3, . . . , an
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