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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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Math Expert
Joined: 02 Sep 2009
Posts: 61283
If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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01 Nov 2019, 05:45
00:00

Difficulty:

85% (hard)

Question Stats:

46% (02:13) correct 54% (02:21) wrong based on 82 sessions

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If $$T= \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \frac{1}{4*5} + ... + \frac{1}{n(n+1)}$$, for some positive integer n, what is the smallest value of n such that $$T>0,97$$?

(A) 29
(B) 30
(C) 31
(D) 32
(E) 33

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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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01 Nov 2019, 05:58
3
2
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

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##### General Discussion
Intern
Joined: 16 Jan 2018
Posts: 43
Location: India
GMAT 1: 620 Q49 V25
GMAT 2: 650 Q49 V28
Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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20 Dec 2019, 12:11
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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Senior Manager
Joined: 21 Jun 2017
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Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
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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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30 Dec 2019, 03:20
1
DEAR moderators,

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

Regards
Intern
Joined: 30 Jan 2017
Posts: 3
Location: India
Concentration: Finance, General Management
WE: Accounting (Accounting)
Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]  [#permalink]

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20 Dec 2019, 11:27
Bunuel could you explain the solution please?
Re: If T = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ......... 1/[n(n+1)]   [#permalink] 20 Dec 2019, 11:27
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