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

Question

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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lacktutor · Accepted Answer

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 Posted from my mobile device

SUNNYRHODE002 · Answer

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

Posted from my mobile device

ShankSouljaBoi · Answer

DEAR moderators,

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

Regards

Kinshook · Answer

Asked: 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 ?

Tn = 1/n(n+1) = 1/n - 1/(n+1)

T = (1/1 - 1/2) + (1/2 - 1/3) + ....+ {1/n - 1/(n+1)} = 1 - 1/(n+1) >.97
1/(n+1) < .03
n+1 > 1/.03 = 100/3 = 33.33
n > 32.33 = 33

IMO E

pgwodehouse · Answer

Can anyone please help in understanding what is wrong with the following solution -

T1 = 1/1*2 = 0.5
T2 = 1/1*2+1/2*3 = 0.84
T3 = 1/1*2+1/2*3+1/3*4 = 0.92
T4 = 1/1*2 + 1/2*3+1/3*4+1/4*3 = 0.97

Hence, n=5.

Kindly help please  nick1816 ,  chetan2u

chetan2u · Answer

You have gone wrong in calculations somewhere. 
1/(1*2)=0.5 and 1/(2*3)=1/6=0.166
So 1/(1*2)+1/(2*3)=0.667

ansarose · Answer

How did the formula in the question suddenly change into the one in the solution with the negative and positive signs..I am confused...could anyone pls explain?

Thekingmaker · Answer

hell yeah i solved this hell it first has to be resolved as 1-1/2 +1/2 - 1/3 ....so on only 1- 1/n will be standing and for it to go beyond 0.97 it had to be least hance 1-1/33 leads to 0.9701 something hence imo E

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If T = 1/(12) + 1/(23) + 1/(34) + 1/(45) + ......... 1/[n(n+1)]

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