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

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

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New post Updated on: 28 Sep 2016, 23:47
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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.

thanks in advance!

Originally posted by RidhimaNigam on 28 Sep 2016, 09:28.
Last edited by Bunuel on 28 Sep 2016, 23:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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New post 28 Sep 2016, 18:32
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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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Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

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New post 28 Sep 2016, 22:58
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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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New post 27 Feb 2017, 06:56
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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New post 27 Feb 2017, 06:56
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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New post 05 Aug 2018, 17:43
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.

thanks in advance!


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

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New post 27 Jan 2019, 15:06
Hello chetan2u !

Could you please explain to me the following rule?

\(x_n=\frac{1}{(n+1)(n+2)}\) as
\(x_n = \frac{1}{n+1} - \frac{1}{n+2}\)

Why can it be written like that?

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

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New post 27 Jan 2019, 17:52
jfranciscocuencag jch103020


Xn = 1/(n+1)(n+2)

[In the numerator, we can write 1 as (n+2)-(n+1) {To recheck, evaluate (n+2)-(n+1) => n+2-n-1 => 1}]

Xn= (n+2) - (n+1)/(n+1)(n+2)
Xn= (n+2)/(n+1)(n+2) - (n+1)/(n+1)(n+2)

Xn=1/(n+1) - 1/(n+2)

Hope it helps :)
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Re: For every positive integer n, if the nth term of the sequence is xN   [#permalink] 27 Jan 2019, 17:52
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