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

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

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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Source: Beat The GMAT — Problem Solving | Thu Sep 13, 2018 11:34 pm

Byjus: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Fri Aug 17, 2018 10:11 pm

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16206; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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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