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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/(

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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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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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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

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=>
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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The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Only $149 for 3 month Online Course
Free Resources-30 day online access & Diagnostic Test
Unlimited Access to over 120 free video lessons-try it yourself
Email to : info@mathrevolution.com

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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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scott@targettestprep.com



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