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If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]

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If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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New post 01 Nov 2019, 06:45
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A
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E

Difficulty:

  85% (hard)

Question Stats:

43% (02:17) correct 57% (02:22) wrong based on 72 sessions

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If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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New post 01 Nov 2019, 06:58
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T= \(1 —\frac{1}{2} + \frac{1}{2} —\frac{1}{3}+ \frac{1}{3} —\frac{1}{4}+ \frac{1}{n} —\frac{1}{(n+1)}= 1–\frac{1}{(n+1)}\)

—> \(1–\frac{1}{(n+1)}> 0.97\)

\(\frac{1}{(n+1)}< \frac{3}{100}\)

\(\frac{1}{(n+1)} —\frac{3}{100} <0\)

\(\frac{(100–3n—3)}{100(n+1)} <0\)

\(\frac{(3n —97)}{100(n+1)} > 0\)

3n—97> 0

\(n > \frac{97}{3}= 32.(3)\)

The smallest value of n is 33

The answer is E

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Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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New post 20 Dec 2019, 12:27
Bunuel could you explain the solution please?
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Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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New post 20 Dec 2019, 13:11
1
Solution :-

1) Split the fractions in the form 1/a -1/b
2) Expand the first 2-3 terms see what happens
3) Only 1/1 and 1/(n+1) will be left and rest will get canceled

Solve further to get values greater than 32.33 .
Hence , Option E

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Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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New post 30 Dec 2019, 04:20
DEAR moderators,

can you please replaye the comma in question to a decimal . It is mentoned as 0,97 .



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Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]   [#permalink] 30 Dec 2019, 04:20
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