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# The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))

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Manager
Joined: 08 Apr 2019
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The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))  [#permalink]

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Updated on: 08 May 2019, 06:54
9
00:00

Difficulty:

95% (hard)

Question Stats:

32% (03:08) correct 68% (03:11) wrong based on 60 sessions

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The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2)). What is the sum of the first 20 terms of the sequence?

(A) 300/440

(B) 325/462

(C) 303/462

(D) 375/450

(E) 650/462

Originally posted by RJ7X0DefiningMyX on 08 May 2019, 06:39.
Last edited by Bunuel on 08 May 2019, 06:54, edited 1 time in total.
Renamed the topic and edited the question.
Director
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The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))  [#permalink]

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08 May 2019, 08:37
1
3
$$\frac{1}{n(n+2)}$$ = $$\frac{1}{2}(\frac{1}{n} - \frac{1}{n+2})$$ ---> (called telescoping series)

so the sum of sequence
= $$\frac{1}{2}(\frac{1}{1} - \frac{1}{3}) + \frac{1}{2}(\frac{1}{2} - \frac{1}{4}) + \frac{1}{2}(\frac{1}{3} - \frac{1}{5}) + \frac{1}{2}(\frac{1}{4} - \frac{1}{6}) + ... + \frac{1}{2}(\frac{1}{18} - \frac{1}{20}) + \frac{1}{2}(\frac{1}{19} - \frac{1}{21}) + \frac{1}{2}(\frac{1}{20} - \frac{1}{22})$$
= $$\frac{1}{2}$$($$\frac{1}{1} + \frac{1}{2} - \frac{1}{21} - \frac{1}{22}) = \frac{1}{2}(\frac{462}{462} + \frac{231}{462} - \frac{22}{462} - \frac{21}{462}) = \frac{1}{2}(\frac{650}{462}) = \frac{325}{462}$$

B
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Re: The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))  [#permalink]

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11 May 2019, 22:39
Mahmoudfawzy83 wrote:
$$\frac{1}{n(n+2)}$$ = $$\frac{1}{2}(\frac{1}{n} - \frac{1}{n+2})$$ ---> (called telescoping series)

so the sum of sequence
= $$\frac{1}{2}(\frac{1}{1} - \frac{1}{3}) + \frac{1}{2}(\frac{1}{2} - \frac{1}{4}) + \frac{1}{2}(\frac{1}{3} - \frac{1}{5}) + \frac{1}{2}(\frac{1}{4} - \frac{1}{6}) + ... + \frac{1}{2}(\frac{1}{18} - \frac{1}{20}) + \frac{1}{2}(\frac{1}{19} - \frac{1}{21}) + \frac{1}{2}(\frac{1}{20} - \frac{1}{22})$$
= $$\frac{1}{2}$$($$\frac{1}{1} + \frac{1}{2} - \frac{1}{21} - \frac{1}{22}) = \frac{1}{2}(\frac{462}{462} + \frac{231}{462} - \frac{22}{462} - \frac{21}{462}) = \frac{1}{2}(\frac{650}{462}) = \frac{325}{462}$$

B

Hi, any other simple approach for this type of questions.. it's too difficult to understand.
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Re: The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))  [#permalink]

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12 May 2019, 01:24
1
mangamma wrote:
Hi, any other simple approach for this type of questions.. it's too difficult to understand.

I don't think it is a good question. it is too hard and technically based on more advanced Calculus theories variations,
so my opinion is to ignore this question.

here are other questions with the same idea but with acceptable level of hardness:
(easy - medium): https://gmatclub.com/forum/if-1-n-n-1-1 ... fl=similar
(hard) : https://gmatclub.com/forum/the-sequence ... l#p2265636 (it is your post mangamma )
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Re: The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))  [#permalink]

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02 Jun 2019, 11:50
Mahmoudfawzy83 wrote:
mangamma wrote:
Hi, any other simple approach for this type of questions.. it's too difficult to understand.

I don't think it is a good question. it is too hard and technically based on more advanced Calculus theories variations,
so my opinion is to ignore this question.

here are other questions with the same idea but with acceptable level of hardness:
(easy - medium): https://gmatclub.com/forum/if-1-n-n-1-1 ... fl=similar
(hard) : https://gmatclub.com/forum/the-sequence ... l#p2265636 (it is your post mangamma )

Thank you
Re: The nth term of a sequence a1, a2, a3 … an is given by an = 1/(n(n+2))   [#permalink] 02 Jun 2019, 11:50
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