For the sequence \(a_1, a_2, a_3 \cdots a_n,\) \(a_n\) is defined by \(a_n=\dfrac1{n}-\dfrac1{n+1}\) for each integer \(n\ge 1.\) What is the sum of the first \(100\) terms of this sequence?
A. \(\dfrac{100}{99}\)
B. \(\dfrac{101}{100}\)
C. \(\dfrac{100}{101}\)
D. \(\dfrac{99}{100}\)
E. \(\dfrac{999}{1010}\)
Answer: C
Source: Veritas Prep
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For the sequence \(a_1, a_2, a_3 \cdots a_n,\) \(a_n\) is defined by \(a_n=\dfrac1{n}-\dfrac1{n+1}\) for each integer
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term1 = 1/1 - 1/2M7MBA wrote: ↑Fri Aug 06, 2021 10:35 amFor the sequence \(a_1, a_2, a_3 \cdots a_n,\) \(a_n\) is defined by \(a_n=\dfrac1{n}-\dfrac1{n+1}\) for each integer \(n\ge 1.\) What is the sum of the first \(100\) terms of this sequence?
A. \(\dfrac{100}{99}\)
B. \(\dfrac{101}{100}\)
C. \(\dfrac{100}{101}\)
D. \(\dfrac{99}{100}\)
E. \(\dfrac{999}{1010}\)
Answer: C
Source: Veritas Prep
term2 = 1/2 - 1/3
term3 = 1/3 - 1/4
term4 = 1/4 - 1/5
.
.
.
term99 = 1/99 - 1/100
term100 = 1/100 - 1/101
So, the sum looks like this:
Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)
NOTICE THAT MOST TERMS CANCEL OUT
Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)
= 1/1 - 1/101
= 101/101 - 1/101
= 100/101
Answer: C
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