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If 1/n(n+1) = 1/n – 1/(n+1), then what is the value of 1/(

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If 1/n(n+1) = 1/n – 1/(n+1), then what is the value of 1/(

by Max@Math Revolution » Thu Sep 13, 2018 11:34 pm

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[Math Revolution GMAT math practice question]

If 1/n(n+1) = 1/n - 1/(n+1), then what is the value of 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)

A. 1/100
B. 1/50
C. 49/50
D. 99/100
E. 1/2
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by Byjus » Fri Sep 14, 2018 1:32 am
Hi,

Given function, 1/n*(n+1) = 1/n - 1/(n+1),

For patterns and sequences questions all we need to do is find initial few values and we will get to understand the pattern to solve the question.

1/(1*2) = 1/1 - ½

1/(2*3) = ½ - 1/3

1/(3*4) = 1/3 - ¼

We can clearly see that, adding the first three terms in the given sum, we will get 1 - ¼ = ¾

Similarly when we add the entire series,

1/1 -1/2 + ½ - 1/3 +......-1/99 + 1/99 -1/00

We will get only first term minus the term left, as all the other terms will get cancelled(subtracted) off

So,

1/1 - 1/100 = 99/100

So the answer is D.

Hope this helps.
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by Brent@GMATPrepNow » Fri Sep 14, 2018 5:52 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If 1/n(n+1) = 1/n - 1/(n+1), then what is the value of 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)

A. 1/100
B. 1/50
C. 49/50
D. 99/100
E. 1/2
If 1/n(n+1) = 1/n - 1/(n+1), then...

1/(1*2) = 1/1 - 1/2
1/(2*3) = 1/2 - 1/3
1/(3*4) = 1/3 - 1/4
.
.
.
1/(98*99) = 1/98 - 1/99
1/(99*100) = 1/99 - 1/100

So, 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100) = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + . . . (1/98 - 1/99) + (1/99 - 1/100)
= 1/1 - 1/100
= 99/100

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Max@Math Revolution » Sun Sep 16, 2018 5:26 pm
=>
1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)
= (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100) = 1/1 - 1/100 = 1 - 1/100 = 99/100 after cancellation of the inner terms.
Therefore, the answer is D.
Answer: D
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by Scott@TargetTestPrep » Tue Sep 18, 2018 5:52 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If 1/n(n+1) = 1/n - 1/(n+1), then what is the value of 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)

A. 1/100
B. 1/50
C. 49/50
D. 99/100
E. 1/2
Using the given formula, we have:

1/(1*2) = 1/1 - 1/2

1/(2*3) = 1/2 - 1/3

1/(3*4) = 1/3 - 1/4

1/(99*100) = 1/99 - 1/100

We see that the required sum is the sum of all the terms of the left hand side of the equal expressions above. Of course, that will be the sum of all the terms of the right hand side also. However, if we add up the terms on the right hand side, we see that we will have the first term of the first equal expressions, 1/1, and the last term of the last equal expressions, -1/100, left (since all the other terms cancel). Therefore, the sum is 1/1 - 1/100 = 99/100.

Answer: D

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