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For every positive integer n, if the nth term of the sequence is
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30 Jan 2020, 00:17
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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 Are You Up For the Challenge: 700 Level Questions
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Re: For every positive integer n, if the nth term of the sequence is
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30 Jan 2020, 00:22
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
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30 Jan 2020, 00:39
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
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For every positive integer n, if the nth term of the sequence is
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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
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30 Jan 2020, 00:58
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
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30 Jan 2020, 01:01
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
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Updated on: 30 Jan 2020, 03:37
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
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30 Jan 2020, 03:37
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
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30 Jan 2020, 03:46
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
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30 Jan 2020, 04:49
\(x_n = \frac{1}{{(n+1)(n+2)}}\)
\(x_1 + x_2 + x_3 + x_4 + x_5 = \frac{1}{{(2)(3)}} + \frac{1}{{(3)(4)}} + \frac{1}{{(4)(5)}} + \frac{1}{{(5)(6)}} +\frac{ 1}{{(6)(7)}}\)
\( = \frac{1}{2}  \frac{1}{3} +\frac{1}{3}  \frac{1}{4} + \frac{1}{4} \frac{1}{5} + \frac{1}{5}  \frac{1}{6} + \frac{1}{6}  \frac{1}{7}\) \(= \frac{1}{2}  \frac{1}{7}\) \( = \frac{(7  2)}{(2*7)}\) \( = \frac{5}{14 }\)



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Re: For every positive integer n, if the nth term of the sequence is
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30 Jan 2020, 05:48
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
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30 Jan 2020, 12:19
For every positive integer n, if the nth term of the sequence is \(x_n=1/(n+1)(n+2)\) then \(x_1+x_2+x_3+x_4+x_5\)=?
> \(\frac{1}{2*3}+ \frac{1}{3*4}+\frac{1}{4*5}+\frac{1}{5*6}+ \frac{1}{6*7} = \)
\(= (\frac{1}{2}\frac{1}{3})+(\frac{1}{3}\frac{1}{4})+(\frac{1}{4}\frac{1}{5})+(\frac{1}{5}\frac{1}{6})+(\frac{1}{6}\frac{1}{7}) =\)
\(= \frac{1}{2} \frac{1}{7}= \frac{(72)}{2*7}= \frac{5}{14}\)
The answer is C.



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For every positive integer n, if the nth term of the sequence is
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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 Archit3110Some 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
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04 Feb 2020, 22:00
Hey! Could someone please explain how is the formula 1/(x+1)(x+2) derived into 1/(x+1)  1/(x2)? Why do we subtract??



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Re: For every positive integer n, if the nth term of the sequence is
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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/(x2)? 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 > 32/(2)(3). That's why there is a subtraction.




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