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The sequence of numbers a1, a2, a3, ..., an is defined by an

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The sequence of numbers a1, a2, a3, ..., an is defined by an  [#permalink]

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New post 09 Jun 2013, 04:31
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The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1. What is the sum of the first 20 terms of this sequence?

A. (1+1/2) - 1/20
B. (1+1/2) - (1/21+1/22)
C. 1 - (1/20 + 1/20)
D. 1 - 1/22
E. 1/20 - 1/22
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Re: The sequence of the number  [#permalink]

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New post 09 Jun 2013, 06:08
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farukqmul wrote:
The sequence of numbers a1, a2, a3, . . . , an, . . . is defined by an= 1/n -1/(n+2)
each integer n>1. What is the sum of the first 20 terms of this sequence?

A (1+1/2) – 1/20
B (1+1/2) –  (1/21 +1/ 22)
C 1 –  (1 /20+ 1/22)
D 1 – 1/22
E 1/20 – 1/22


The answer would most certainly be [B]. But the question needs a slight modification. n>=1, since the answer does consider a1 under the sum.

The sequence is :

a1 = 1-1/3
a2 = 1/2 - 1/4
a3 = 1/3 - 1/5....

We can observe that the third term in the sequence cancels the negative term in the first. A similar approach can be seen on all the terms and we would be left with 1 + 1/2 from a1 and a2 along with -1/22 and -1/21 from a20 and a19 term which could not be cancelled.

Hence the sum = (1+1/2) –  (1/21 +1/ 22). Hope it helps!

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Re: The sequence of numbers a1, a2, a3, ..., an is defined by an  [#permalink]

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New post 09 Jun 2013, 07:00
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farukqmul wrote:
The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1. What is the sum of the first 20 terms of this sequence?

A. (1+1/2) - 1/20
B. (1+1/2) - (1/21+1/22)
C. 1 - (1/20 + 1/20)
D. 1 - 1/22
E. 1/20 - 1/22


\(a_1+a_2+a_3+.a_4..+a_{18}+a_{19}+a_{20}=(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{18}-\frac{1}{20})+(\frac{1}{19}-\frac{1}{21})+(\frac{1}{20}-\frac{1}{22})=1+\frac{1}{2}-\frac{1}{21}-\frac{1}{22}\) (all numbers except these 4 cancel).

Answer: B.
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Re: The sequence of numbers a1, a2, a3, ..., an is defined by an  [#permalink]

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New post 28 Sep 2016, 06:18
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farukqmul wrote:
The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1. What is the sum of the first 20 terms of this sequence?

A. (1+1/2) - 1/20
B. (1+1/2) - (1/21+1/22)
C. 1 - (1/20 + 1/20)
D. 1 - 1/22
E. 1/20 - 1/22


this is a type of the questions that i hate the most...
i got to the solution by applying logic rather than crunching with the numbers...

fact 1. the sum will be >1. threfore, C, D, and E are out.
fact 2. since we have 1/(n+2), and last term for n is 20, then there must be smth 1/22. A is out, and B stands.
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Re: The sequence of numbers a1, a2, a3, ..., an is defined by an  [#permalink]

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New post 01 Mar 2017, 02:32
We can see that the summation will be 1-⅓ +½-¼ +⅓-⅕ + ……..1/18 -1/20 +1/19 -1/21 +1/20 - 1/22
After cancellation we get 1 +½ -1/21-1/22. Option B
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Re: The sequence of numbers a1, a2, a3, ..., an is defined by an  [#permalink]

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Re: The sequence of numbers a1, a2, a3, ..., an is defined by an &nbs [#permalink] 22 Aug 2018, 00:10
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