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B
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65% (01:46) correct
35%
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What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)
A. \(\frac{98}{99}\)
B. \(\frac{99}{100}\)
C. \(\frac{100}{101}\)
D. \(\frac{98}{101}\)
E. \(\frac{99}{101}\)
A. \(\frac{98}{99}\)
B. \(\frac{99}{100}\)
C. \(\frac{100}{101}\)
D. \(\frac{98}{101}\)
E. \(\frac{99}{101}\)
Updated on: 26 Sep 2020, 07:33
Originally posted by BrentGMATPrepNow on 28 Nov 2016, 08:05.
Last edited by BrentGMATPrepNow on 26 Sep 2020, 07:33, edited 1 time in total.
Last edited by BrentGMATPrepNow on 26 Sep 2020, 07:33, edited 1 time in total.
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nguyendinhtuong
What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)
A. \(\frac{98}{99}\)
B. \(\frac{99}{100}\)
C. \(\frac{100}{101}\)
D. \(\frac{98}{101}\)
E. \(\frac{99}{101}\)
A. \(\frac{98}{99}\)
B. \(\frac{99}{100}\)
C. \(\frac{100}{101}\)
D. \(\frac{98}{101}\)
E. \(\frac{99}{101}\)
Let's find the sum of various fractions and look for a pattern
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}\) = 1/2 + 1/6
= 3/6 + 1/6
= 4/6
= 2/3
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) = 2/3 + \(\frac{1}{3 \times 4}\)
= 2/3 + 1/12
= 8/12 + 1/12
= 9/12
= 3/4
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\frac{1}{4 \times 5}\) = 3/4 + \(\frac{1}{4 \times 5}\)
= 3/4 + 1/20
= 15/20 + 1/20
= 16/2
= 4/5
As we can see, the pattern is...
\(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{(n) \times (n+1)}\) = \(\frac{n}{n+1}\)
So, \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\) = \(\frac{99}{100}\)
Answer: B
29 Nov 2016, 15:53
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nguyendinhtuong
What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)
A. \(\frac{98}{99}\)
B. \(\frac{99}{100}\)
C. \(\frac{100}{101}\)
D. \(\frac{98}{101}\)
E. \(\frac{99}{101}\)
A. \(\frac{98}{99}\)
B. \(\frac{99}{100}\)
C. \(\frac{100}{101}\)
D. \(\frac{98}{101}\)
E. \(\frac{99}{101}\)
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
