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

Question

Competition Mode Question

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

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GMATinsight · Answer

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

rajatchopra1994 · Answer

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

Archit3110 · Answer

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

Staphyk · Answer

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

Posted from my mobile device

freedom128 · Answer

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)

Posted from my mobile device

exc4libur · Answer

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*3+1/3*4=2+4/2*3*4=1/4
  1/4*5+1/5*6=6+4/4*5*6=10/120=1/12
  1/4+1/12=3+1/4*3=4/12=1/3
 =1/3+1/6*7=1/3+1/3*14=15/42=5/14

Ans (C)

CareerGeek · Answer

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

eakabuah · Answer

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 = /420 = 150/420 = 15/42 = 5/14

The answer is C.

Posted from my mobile device

hardikajmani · Answer

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 }

shameekv1989 · Answer

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

lacktutor · Answer

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.

GMATinsight · Answer

Archit3110 Some typo in highlighted part ... should be 1/42

IevaPieva · Answer

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??

diegoahernandezb · Answer

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.

hasanloubani · Answer

@

may i know how did you get the red part? this is the most efficient way of solving the question, but i didn't get the 150/420.

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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

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