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Jan 05, 2021 6:15 am
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D
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Difficulty:
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Question Stats:
72% (01:33) correct
28%
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wrong
based on 451
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If \(F(X) = \frac{1}{X} - \frac{1}{X+1}\), what is the sum of F(1) through F(200), inclusive?
A. 0
B. 1
C. 100
D. 200/201
E. 1/40200
A. 0
B. 1
C. 100
D. 200/201
E. 1/40200
Jan 05, 2021 7:21 pm
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Bunuel
From the given information, we get...
\(f(1) = 1 - \frac{1}{2}\)
\(f(2) = \frac{1}{2} - \frac{1}{3}\)
\(f(3) = \frac{1}{3} - \frac{1}{4}\)
\(f(4) = \frac{1}{4} - \frac{1}{5}\)
.
.
.
\(f(199) = \frac{1}{199} - \frac{1}{200}\)
\(f(200) = \frac{1}{200} - \frac{1}{201}\)
So, \(f(1) + f(2) + f(3) + f(4) + . . . . f(199) + f(200) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) . . . . . + (\frac{1}{199} - \frac{1}{200}) + (\frac{1}{200} - \frac{1}{201}) \)
Notice that almost every POSITIVE fraction has an identical NEGATIVE fraction.
So all of those fractions add up to be ZERO
The only two values that don't cancel out are the first and last terms (1 and -1/201)
In other words, the sum \(= 1 - \frac{1}{201} = \frac{201}{201}-\frac{1}{201}=\frac{200}{201}\)
Answer: D
General Discussion
XSatishX
Joined: 06 Jul 2020
Last visit: 18 Nov 2022
Posts: 100
Given Kudos: 36
Location: India
Concentration: Leadership, General Management
Schools: LBS '24 (D) ISB '23 (II) Tepper '24 (S)
GMAT 1: 650 Q49 V30
GMAT 2: 730 Q48 V41 (Online)
GPA: 3.8
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Jan 05, 2021 6:25 am
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F(x) = 1/x -1/(x+1)
F(1) = 1-1/2
F(2)=1/2 - 1/3
F(3) =1/3-1/4
Now if we see the sequence here, Every second term of previous number will cancel the first term of following number.
For eg: F(1)+F(2)+F(3) = 1-1/2+1/2-1/3+1/3-1/4 = 1-1/4 or 1/starting number - (1/end term+1)
So Sum of all numbers through 1 to 200 will be
F(1)+F(2)+F(3)....F(200) = 1/1 -1/(200+1)
=1-1/201 = 200/201
Answer D
F(1) = 1-1/2
F(2)=1/2 - 1/3
F(3) =1/3-1/4
Now if we see the sequence here, Every second term of previous number will cancel the first term of following number.
For eg: F(1)+F(2)+F(3) = 1-1/2+1/2-1/3+1/3-1/4 = 1-1/4 or 1/starting number - (1/end term+1)
So Sum of all numbers through 1 to 200 will be
F(1)+F(2)+F(3)....F(200) = 1/1 -1/(200+1)
=1-1/201 = 200/201
Answer D
Jun 29, 2024 4:30 am
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
\(F(X) = \frac{1}{X} - \frac{1}{X+1}\) and we need to find the value of F(1) + F(2) + .... + F(200)
F(1) = \(\frac{1}{X} - \frac{1}{X+1}\) = \(\frac{1}{1} - \frac{1}{1+1}\) = \(\frac{1}{1} - \frac{1}{2}\)
=> F(1) + F(2) + .... + F(200) = (\(\frac{1}{1} - \frac{1}{2}\)) + (\(\frac{1}{2} - \frac{1}{3}\)) + (\(\frac{1}{3} - \frac{1}{4}\)) + .... + (\(\frac{1}{199} - \frac{1}{200}\)) + (\(\frac{1}{200} - \frac{1}{201}\))
Alternate terms will cancel out and we will be left with only the first and the last term
=> F(1) + F(2) + .... + F(200) = \(\frac{1}{1} - \frac{1}{201}\) = \(\frac{201 - 1}{201}\) = \(\frac{200}{201}\)
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Functions and Custom Characters
\(F(X) = \frac{1}{X} - \frac{1}{X+1}\) and we need to find the value of F(1) + F(2) + .... + F(200)
F(1) = \(\frac{1}{X} - \frac{1}{X+1}\) = \(\frac{1}{1} - \frac{1}{1+1}\) = \(\frac{1}{1} - \frac{1}{2}\)
=> F(1) + F(2) + .... + F(200) = (\(\frac{1}{1} - \frac{1}{2}\)) + (\(\frac{1}{2} - \frac{1}{3}\)) + (\(\frac{1}{3} - \frac{1}{4}\)) + .... + (\(\frac{1}{199} - \frac{1}{200}\)) + (\(\frac{1}{200} - \frac{1}{201}\))
Alternate terms will cancel out and we will be left with only the first and the last term
=> F(1) + F(2) + .... + F(200) = \(\frac{1}{1} - \frac{1}{201}\) = \(\frac{201 - 1}{201}\) = \(\frac{200}{201}\)
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Functions and Custom Characters
