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Updated on: 18 Jun 2023, 04:26
Originally posted by inakihernandez on 24 Apr 2016, 10:38.
Last edited by Bunuel on 18 Jun 2023, 04:26, edited 2 times in total.
Last edited by Bunuel on 18 Jun 2023, 04:26, edited 2 times in total.
Edited the question.
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E
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For every positive integer n, the \(n_{th}\) term of sequence is given by \(a_n = \frac{1}{n} - \frac{1}{n+1}\). What is the sum of the first 100 terms?
(A) 1
(B) 0
(C) 25
(D) 99/100
(E) 100/101
(A) 1
(B) 0
(C) 25
(D) 99/100
(E) 100/101
24 Apr 2016, 10:47
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inakihernandez
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/n+1. What is the sum of the first 100 terms?
(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
Hi,
the moment you see the formula of nth number, it should tell you we are looking at a sequence ..
\(a_1 = \frac{1}{1}-\frac{1}{2}\)..
\(a_2= \frac{1}{2}-\frac{1}{3}\)..
\(a_3 = \frac{1}{3}-\frac{1}{4}\)..
.....
\(a_{99} = \frac{1}{99}-\frac{1}{100}\)..
\(a_{100} = \frac{1}{100}- \frac{1}{101}\)..
ADD all
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101} = 1-\frac{1}{101} = \frac{100}{101}\)
so when you add all terms, what is left is \(1-\frac{1}{101} = \frac{100}{101}\)..
E
28 Apr 2018, 07:11
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inakihernandez
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?
(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
term1 = 1/1 - 1/2
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
= E
Cheers,
Brent
