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Sequence

Expert replies
by MBA.Aspirant » Sun Jun 26, 2011 5:43 pm
The sequence of numbers a1,a2,a3.....an is defined by an = 1/n - 1/n+2 for each integer n >or= 1. What is the sum of the first 20 terms of this sequence?

A) (1+1/2) - 1/20
B) (1+1/2) - (1/21+1/22)
C) 1 - (1/20 + 1/20)
D) 1 - 1/22
E) 1/20 - 1/22
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Source: — Problem Solving |

by Frankenstein » Sun Jun 26, 2011 9:13 pm
Hi,
t1 = 1... -1/3
t2 = ..1/2.....-1/4
t3 = ......1/3......-1/5
t4 = ............1/4......-1/6
.
.
.
t18 = ...................................... 1/18....-1/20
t19 = ...........................................1/19......-1/21
t20 = .................................................1/20.......-1/22
------------------------------------------------------------------------
Sum = 1+1/2+0+0+0...................................+0..-1/21 -1/22
So, sum = (1+1/2) - (1/21+1/22)

Hence, B
Cheers!

Things are not what they appear to be... nor are they otherwise
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by MBA.Aspirant » Sun Jun 26, 2011 11:40 pm
Thanks Frankenstein. Should you be doing the calculations or just cancelling out the fractions with each other? I think you have to lay them out all first then cancel. But this will take much time...any faster approach?
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by Frankenstein » Mon Jun 27, 2011 12:29 am
MBA.Aspirant wrote:Thanks Frankenstein. Should you be doing the calculations or just cancelling out the fractions with each other? I think you have to lay them out all first then cancel. But this will take much time...any faster approach?
Hi,
Just cancelling. No calculations. If you want faster approach, we can perhaps write in a different manner even though both are same.
write all values of 1/n as 1,1/2,...1/20
then write all values of 1(n+2) as 1/3,1/4,...1,22
While subtracting, we can see that common elements from 1/3 to 1/20 are cancelled and we are left with 1+1/2 from first series and (1/21+1/22) from the second series.
Cheers!

Things are not what they appear to be... nor are they otherwise
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