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For every positive integer n, if the nth term of the sequence is xN

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Joined: 29 Jul 2016
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For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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Updated on: 28 Sep 2016, 22:47
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Difficulty:

45% (medium)

Question Stats:

70% (02:00) correct 30% (01:50) wrong based on 90 sessions

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For every positive integer n, if the nth term of the sequence is $$x_n=\frac{1}{(n+1)(n+2)}$$ then $$x_1+x_2+x_3+x_4+x_5=$$?

a) 4/5
b) 2/3
c) 5/14
d) 5/16
e) 6/19

Could anyone explain me how to do this without the normal convention of adding the 5 terms.

Originally posted by RidhimaNigam on 28 Sep 2016, 08:28.
Last edited by Bunuel on 28 Sep 2016, 22:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Joined: 05 Jun 2015
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WE: Engineering (Transportation)
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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28 Sep 2016, 17:32
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Write $$x_n=\frac{1}{(n+1)(n+2)}$$ as
$$x_n = \frac{1}{n+1} - \frac{1}{n+2}$$

$$x_1+x_2 = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4})$$
= $$\frac{1}{2} - \frac{1}{4}$$

Only the first term and the last term remain when we add multiple $$x_n$$ values.

$$x_5 = \frac{1}{6} - \frac{1}{7}$$
Therefore we have $$x_1+x_2+x_3+x_4+x_5= \frac{1}{2} - \frac{1}{7}$$
=$$\frac{5}{14}$$
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Posts: 5
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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28 Sep 2016, 21:58
1
rudra316 wrote:
Write $$x_n=\frac{1}{(n+1)(n+2)}$$ as
$$x_n = \frac{1}{n+1} - \frac{1}{n+2}$$

$$x_1+x_2 = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4})$$
= $$\frac{1}{2} - \frac{1}{4}$$

Only the first term and the last term remain when we add multiple $$x_n$$ values.

$$x_5 = \frac{1}{6} - \frac{1}{7}$$
Therefore we have $$x_1+x_2+x_3+x_4+x_5= \frac{1}{2} - \frac{1}{7}$$
=$$\frac{5}{14}$$

Oh yeah! Thanks

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Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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27 Feb 2017, 05:56
1/(n+1)(n+2) can be written as 1/(n+1) - 1/(n+2)
Therefore x1 = ½ -⅓
x2 = ⅓ -¼ and so in.
Therefore x1+x2+x3+x4+x5 = ½-⅓ +⅓-¼ +¼ -⅕ +⅕ -⅙ +⅙ - 1/7 =½ - 1/7 = 5/14
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Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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27 Feb 2017, 05:56
1/(n+1)(n+2) can be written as 1/(n+1) - 1/(n+2)
Therefore x1 = ½ -⅓
x2 = ⅓ -¼ and so on.
Therefore x1+x2+x3+x4+x5 = ½-⅓ +⅓-¼ +¼ -⅕ +⅕ -⅙ +⅙ - 1/7 =½ - 1/7 = 5/14
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Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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05 Aug 2018, 16:43
RidhimaNigam wrote:
For every positive integer n, if the nth term of the sequence is $$x_n=\frac{1}{(n+1)(n+2)}$$ then $$x_1+x_2+x_3+x_4+x_5=$$?

a) 4/5
b) 2/3
c) 5/14
d) 5/16
e) 6/19

Could anyone explain me how to do this without the normal convention of adding the 5 terms.

I must be missing something basic here, but what is the rule to rewrite the equation as below:

Write xn=1(n+1)(n+2) as
xn=1n+1−1n+2
Re: For every positive integer n, if the nth term of the sequence is xN &nbs [#permalink] 05 Aug 2018, 16:43
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