What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?

Question

What is the value of  A=\frac{1}{1 	imes 2}+\frac{1}{2 	imes 3}+\frac{1}{3 	imes 4}+...+\frac{1}{99 	imes 100}

A.  \frac{98}{99}

B.  \frac{99}{100}

C.  \frac{100}{101}

D.  \frac{98}{101}

E.  \frac{99}{101}

BrentGMATPrepNow · Accepted Answer

Let's find the sum of various fractions and  look for a pattern

\frac{1}{1 	imes 2}+\frac{1}{2 	imes 3}  = 1/2 + 1/6
= 3/6 + 1/6
= 4/6
=  2/3

\frac{1}{1 	imes 2}+\frac{1}{2 	imes 3}+\frac{1}{3 	imes 4}  =  2/3  +  \frac{1}{3 	imes 4}  
= 2/3 + 1/12
= 8/12 + 1/12
= 9/12
=  3/4

\frac{1}{1 	imes 2}+\frac{1}{2 	imes 3}+\frac{1}{3 	imes 4}+\frac{1}{4 	imes 5}  =  3/4  +  \frac{1}{4 	imes 5} 
= 3/4 + 1/20
= 15/20 + 1/20
= 16/2
=  4/5

As we can see, the pattern is...
 A=\frac{1}{1 	imes 2}+\frac{1}{2 	imes 3}+\frac{1}{3 	imes 4}+...+\frac{1}{(n) 	imes (n+1)}  =  \frac{n}{n+1}

So,  \frac{1}{1 	imes 2}+\frac{1}{2 	imes 3}+\frac{1}{3 	imes 4}+...+\frac{1}{99 	imes 100}  =  \frac{99}{100}

Answer: B

ScottTargetTestPrep · Answer

We begin by noting that 1/(1x2) = 1/1 - 1/2, 1/(2x3) = 1/2 - 1/3, 1/(3x4) = 1/3 - 1/4, … , 1/(99x100) = 1/99 - 1/100.

If we substitute each term in the definition of A by the above, we obtain:

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + … + 1/99 - 1/100

Note that all of the terms (besides 1/1 = 1 and 1/100) cancel; therefore,

A = 1/1 - 1/100 = 99/100

Answer: B

Bunuel · Answer

Similar question to practice: what-is-168951.html

sjavvadi · Answer

Please give Kudos+1 if you like the method

VS93 · Answer

Although I can follow the explanations made in this thread I would not be able to come up with them on my own. How can we make sure that we identify the right pattern in such a question?

sjavvadi · Answer

In general when I see a question with summation of products (numerator or denominator) I try to see if I can express the general term as a difference. This would help us cancel the terms in a summation series...

Sent from my Nexus 6 using  GMAT Club Forum mobile app

Cez005 · Answer

n=1: 1/2
n=2: 1/2 + 1/2x3 = 1/2 + 1/6 = 2/3
n=3: 4/6 + 1/3x4 = 4/6 + 1/12 = 3/4
n=4: 3/4 + 1/4x5 = 3/4 + 1/20 = 4/5

So, we can deduce that: n(x) = (x-1/x) + (1/x(x+1))

Therefore, n(99) = (98/99) + (1/99*100) = 9801/9900 = 99/100

Alphatango · Answer

No need to use pen /pencil
Ex :
1/(1*2) + 1/(2*3)=2/3
Similarly if u expand it you will find 3/4,4/5 going through.
Hence ans last digit 99/100

hanyhamdani · Answer

Best way to solve these kind of questions is to simplify them to a known form.
1/2 + 1/(2x3) + 1/(3x4) + .............+ 1/(99x100)

since 1/(2x3)= 1/6 = 1/2-1/3, In a similar way all terms can be written in to this form
=1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4- 1/5+............-1/99+1/99-1/100
=1/2 + 1/2 -1/100 (all others cancelled)
=1-1/100 = 99/100
B

Thelegend2631 · Answer

i

Keep practicing similar q's. you will get there some day.

KarishmaB · Answer

Questions with this many terms are bound to use logic that cancel off most terms.

The denominators have 1*2, 2*3, 3*4 etc in them so try to split each term into one positive and one negative term.

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

Adding all left hand sides will cancel off all terms except first and last 1 - 1/100 = 99/100

99/100 = 1/1*2 + 1/2*3 + 1/3*4 ... + 1/99*100

Answer (B)

Crytiocanalyst · Answer

One key aspect is to undestand thaat 
1/(1x2) = 1/1 - 1/2,
1/(2x3) = 1/2 - 1/3
 1/(3x4) = 1/3 - 1/4, … , 
1/(99x100) = 1/99 - 1/100.

And then we solve for the same we get the answer

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + … + 1/99 - 1/100

only term that's left is 1-1/100

A = 1/1 - 1/100 = 99/10
Hence IMO B

bumpbot · Answer

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