[GMAT math practice question]
What is the value of 2(1-1/2) + 3(1-1/3) + ... + 100(1 - 1/100)?
A. 3600
B. 4800
C. 4950
D. 5000
E. 5050
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What is the value of 2(1-1/2) + 3(1-1/3) + … + 100(1 – 1
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Max@Math Revolution
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C
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E
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Solve the brackets and you get:-
2*1/2+3*2/3+...... 100*99/100
cancelling out all the numbers we get:-
1+2+3+4...........99
Apply the following formula:-
n/2 (a1+an)
99/2 (1+99)
Sum= 4950 $$$$ [/list]
2*1/2+3*2/3+...... 100*99/100
cancelling out all the numbers we get:-
1+2+3+4...........99
Apply the following formula:-
n/2 (a1+an)
99/2 (1+99)
Sum= 4950 $$$$ [/list]
Last edited by pkithania on Sun Mar 17, 2019 10:13 pm, edited 1 time in total.
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Max@Math Revolution
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=>
Note: This sum has 100 - 2 + 1 = 99 terms.
2(1-1/2) + 3(1-1/3) + ... + 100(1 - 1/100)
= (2 - 1) + (3 - 1) + ... + (100 - 1)
= 2 + 3 + ... + 100 - ( 1 + 1 + ... + 1 )
= ( 2 + 100 )99 / 2 - 99
= 51*99 - 99 = (51-1)*99 = 50*99 = 4950
Therefore, the answer is C.
Answer: C
Note: This sum has 100 - 2 + 1 = 99 terms.
2(1-1/2) + 3(1-1/3) + ... + 100(1 - 1/100)
= (2 - 1) + (3 - 1) + ... + (100 - 1)
= 2 + 3 + ... + 100 - ( 1 + 1 + ... + 1 )
= ( 2 + 100 )99 / 2 - 99
= 51*99 - 99 = (51-1)*99 = 50*99 = 4950
Therefore, the answer is C.
Answer: C
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Scott@TargetTestPrep
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Simplifying we see that we have:Max@Math Revolution wrote:[GMAT math practice question]
What is the value of 2(1-1/2) + 3(1-1/3) + ... + 100(1 - 1/100)?
A. 3600
B. 4800
C. 4950
D. 5000
E. 5050
(2 - 1) + (3 - 1) + ... + (100 - 1) = 1 + 2 + .... + 99
So we see that we need to determine the sum of the consecutive integers from 1 to 99, inclusive:
sum = average x number
sum = (1 + 99)/2 x 99
sum = 50 x 99 = 4950
Answer: C
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