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

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

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New post 30 Jan 2020, 00:17
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 30 Jan 2020, 00:22
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Quote:
For every positive integer n, if the nth term of the sequence is x_n=1/(n+1)(n+2) then x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19


\(x_1 = 1/(2*3) = (1/2)-(1/3)\)
\(x_2 = 1/(3*4) = (1/3)-(1/4)\)
\(x_3 = 1/(4*5) = (1/4)-(1/5)\)
\(x_4 = 1/(5*6) = (1/5)-(1/6)\)
\(x_5 = 1/(6*7) = (1/6)-(1/7)\)

\(x_1 + x_2 + x_3 + x_4 + x_5 = (1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+(1/5)-(1/6)+(1/6)-(1/7) = (1/2)-(1/7) = 5/14\)

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

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New post 30 Jan 2020, 00:39
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Explanation:

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

So,
X1 = 1/2 - 1/3
X2 = 1/3 - 1/4
X3 = 1/4 - 1/5
X4 = 1/5 - 1/6
X5 = 1/6 - 1/7

x1+x2+x3+x4+x5 = 1/2- 1/7 = 5/14

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

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New post Updated on: 31 Jan 2020, 09:22
for the given function ; find x1 to x5
we get x1 = 1/6 x2= 1/12 x3= 1/20 x4= 1/30 x5= 1/42
take LCM
sum from x1 to x5 = 150/420 ; 5/14
IMO C


For every positive integer n, if the nth term of the sequence is xn=1(n+1)(n+2)xn=1(n+1)(n+2) then x1+x2+x3+x4+x5=x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Originally posted by Archit3110 on 30 Jan 2020, 00:51.
Last edited by Archit3110 on 31 Jan 2020, 09:22, edited 1 time in total.
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 30 Jan 2020, 00:58
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For every positive integer n, if the nth term of the sequence is
Xn = 1/(n+1)(n+2) then X1+X2+X3+X4+X5 =?

1/2(3)+1/3(4)+ 1/4(5) + 1/5(6)+1/6(7)
1/6 + 1/12+ 1/20 + 1/30 + 1/42
3/12+ 1/20 +1/30+ 1/42
15/60+ 3/60 + 2/60 +1/42
1/3 + 1/42 —> 14/42 + 1/42 = 5/14

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Hit that C

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

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New post 30 Jan 2020, 01:01
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x1+x2+x3+x4+x5
=1/6+1/12+1/20+1/30+1/42
=1/420*(70+35+21+14+10)
=150/420
=5/14

FINAL ANSWER IS (C)

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

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New post Updated on: 30 Jan 2020, 03:37
1
Quote:
For every positive integer n, if the nth term of the sequence is x_n=1/(n+1)(n+2) then x1+x2+x3+x4+x5= ?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19


x_n=1/(n+1)(n+2)
x_1=1/2*3, x_2=1/3*4, x_3=1/4*5, x_4=1/5*6, x_5=1/6*7
lcm(3,4,5,7)=420: sum(x_1 to x_5)=(70+35+21+14+10)/420=5/14

OR
[1+2] 1/2*3+1/3*4=2+4/2*3*4=1/4
[3+4] 1/4*5+1/5*6=6+4/4*5*6=10/120=1/12
[1,2,3,4] 1/4+1/12=3+1/4*3=4/12=1/3
[(1,2,3,4)+5]=1/3+1/6*7=1/3+1/3*14=15/42=5/14

Ans (C)

Originally posted by exc4libur on 30 Jan 2020, 03:30.
Last edited by exc4libur on 30 Jan 2020, 03:37, edited 1 time in total.
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 30 Jan 2020, 03:37
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Given, \(x_n = \frac{1}{(n+1)(n+2)}\)
--> \(x_1\) + \(x_2\) + \(x_3\) + \(x_4\) + \(x_5\) = \(\frac{1}{(2)(3)} + \frac{1}{(3)(4)} + \frac{1}{(4)(5)} + \frac{1}{(5)(6)} + \frac{1}{(6)(7)}\)
= \(\frac{3 - 2}{2*3} + \frac{4 - 3}{3*4} + \frac{5 - 4}{4*5} + \frac{6 - 5}{5*6} + \frac{7 - 6}{6*7}\)
= \(\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}\)
= \(\frac{1}{2} - \frac{1}{7}\)
= \(\frac{7 - 2}{2*7}\)
= \(\frac{5}{14}\)

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

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New post 30 Jan 2020, 03:46
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x1=1/(2*3)=1/6
x2=1/(3*4)=1/12
x3=1/(4*5)=1/20
x4=1/(5*6)=1/30
x5=1/(6*7)=1/42

x1+x2+x3+x4+x5 = 1/6+1/12+1/20+1/30+1/42 =[70+35+21+14+10]/420 = 150/420 = 15/42 = 5/14

The answer is C.

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

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New post 30 Jan 2020, 04:49
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\(x_n = \frac{1}{{(n+1)(n+2)}}\)

\(x_1 + x_2 + x_3 + x_4 + x_5 = \frac{1}{{(2)(3)}} + \frac{1}{{(3)(4)}} + \frac{1}{{(4)(5)}} + \frac{1}{{(5)(6)}} +\frac{ 1}{{(6)(7)}}\)

\( = \frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4} + \frac{1}{4} -\frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}\)
\(= \frac{1}{2} - \frac{1}{7}\)
\( = \frac{(7 - 2)}{(2*7)}\)
\( = \frac{5}{14 }\)
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 30 Jan 2020, 05:48
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For every positive integer n, if the nth term of the sequence is xn=1(n+1)(n+2)xn=1(n+1)(n+2) then x1+x2+x3+x4+x5=x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

x1 = \(1/(2*3)\)
x2 = \(1/(3*4)\)

x1 + x2 = \(1/4\)

x3 = \(1/(4*5)\)
x4 = \(1/(5*6)\)

x3 + x4 = \(1/12\)

x5 = \(1/42\)

Therefore x1 + x2 + x3 + x4 + x5 = \(1/4+1/12+1/42\) -> \((21+7+2)/84\) -> \(5/14\) - Answer C
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 30 Jan 2020, 12:19
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For every positive integer n, if the nth term of the sequence is \(x_n=1/(n+1)(n+2)\) then \(x_1+x_2+x_3+x_4+x_5\)=?

--> \(\frac{1}{2*3}+ \frac{1}{3*4}+\frac{1}{4*5}+\frac{1}{5*6}+ \frac{1}{6*7} = \)

\(= (\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7}) =\)

\(= \frac{1}{2} -\frac{1}{7}= \frac{(7-2)}{2*7}= \frac{5}{14}\)

The answer is C.
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For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 30 Jan 2020, 23:56
Archit3110 wrote:
for the given function ; find x1 to x5
we get x1 = 1/6 x2= 1/12 x3= 1/20 x4= 1/30 x5= 1/58
sum from x1 to x5 = 75/240 ; 5/16
IMO D


For every positive integer n, if the nth term of the sequence is xn=1(n+1)(n+2)xn=1(n+1)(n+2) then x1+x2+x3+x4+x5=x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Archit3110
Some typo in highlighted part ... should be 1/42 :)
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 04 Feb 2020, 22:00
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Hey! Could someone please explain how is the formula 1/(x+1)(x+2) derived into 1/(x+1) - 1/(x-2)? Why do we subtract??
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Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

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New post 21 Feb 2020, 11:14
IevaPieva wrote:
Hey! Could someone please explain how is the formula 1/(x+1)(x+2) derived into 1/(x+1) - 1/(x-2)? Why do we subtract??


I was dealing with the same question to improve the way in which this type of question is solved. Although I cannot explain the theory behind this reasoning, I would try to show an example:

We have that --> xn = 1/(x+1)(x+2) and therefore, for x1 = 1 / (1+1)(1+2) --> 1 / (2)(3) --> 1/6.

It turns that another way to obtain 1/6 is if we separate both fractions (as in the solution provided by the other members). x1 = 1 / (1+1) - 1 /(1+2) --> 1/2 - 1/3 --> 3-2/(2)(3). That's why there is a subtraction.
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Re: For every positive integer n, if the nth term of the sequence is   [#permalink] 21 Feb 2020, 11:14
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