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For every positive integer n, the nth term of a sequence is given by
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18 Jun 2023, 04:22
18 Jun 2023, 04:22
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For every positive integer n, the nth term of a sequence is given by \(a_n=\frac{1}{n} - \frac{1}{n+1}\). What is the sum of the first 100 terms of this sequence?
(A) 0
(B) 1/101
(C) 99/100
(D) 100/101
(E) 1
QR31.png [ 33.42 KiB | Viewed 3201 times ]
(A) 0
(B) 1/101
(C) 99/100
(D) 100/101
(E) 1
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QR31.png [ 33.42 KiB | Viewed 3201 times ]
Note: This question appears with different answer choices in another GMAT Prep question.
Re: For every positive integer n, the nth term of a sequence is given by
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18 Jun 2023, 04:34
18 Jun 2023, 04:34
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a1 = (1/1)-(1/2)
a2 = (1/2)-(1/3)
...
a99 = (1/99)-(1/100)
a100 = (1/100)-(1/101)
s100 = a1+a2+......+a99+a100
= (1/1)-(1/2)+(1/2)-(1/3)+(1/99)-(1/100)+(1/100)-(1/101)
= 1-1/101
= 100/101
Posted from my mobile device
a2 = (1/2)-(1/3)
...
a99 = (1/99)-(1/100)
a100 = (1/100)-(1/101)
s100 = a1+a2+......+a99+a100
= (1/1)-(1/2)+(1/2)-(1/3)+(1/99)-(1/100)+(1/100)-(1/101)
= 1-1/101
= 100/101
Posted from my mobile device
For every positive integer n, the nth term of a sequence is given by
[#permalink]
18 Jun 2023, 05:22
18 Jun 2023, 05:22
1
Kudos
Quote:
For every positive integer n, the nth term of a sequence is given by \(a_n=\frac{1}{n} - \frac{1}{n+1}\). What is the sum of the first 100 terms of this sequence?
\(a_1=\frac{{1}}{{1}}-\frac{{1}}{{2}}\)
\(a_2=\frac{{1}}{{2}}-\frac{{1}}{{3}}\)
\(a_3=\frac{{1}}{{3}}-\frac{{1}}{{4}}\)
\(a_4=\frac{{1}}{{4}}-\frac{{1}}{{5}}\)
\(a_5=\frac{{1}}{{5}}-\frac{{1}}{{6}}\)
.
.
.
\(a_{99}=\frac{{1}}{{99}}-\frac{{1}}{{100}}\)
\(a_{100}=\frac{{1}}{{100}}-\frac{{1}}{{101}}\)
Hence, adding and cancelling all the like terms (\( \pm \));
\(S_{100}=\frac{{1}}{{1}}-\frac{{1}}{{101}}=\frac{{100}}{{101}}\) Option (D)
