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# What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?

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What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?  [#permalink]

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28 Nov 2016, 03:59
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Difficulty:

95% (hard)

Question Stats:

30% (02:16) correct 70% (02:15) wrong based on 69 sessions

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What is the value of $$A=\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+...+\frac{1}{48 \times 49 \times 50}$$?

A. $$\frac{306}{1225}$$

B. $$\frac{637}{2550}$$

C. $$\frac{1175}{4704}$$

D. $$\frac{1199}{4800}$$

E. $$\frac{2497}{9996}$$

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What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?  [#permalink]

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28 Nov 2016, 07:48
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nguyendinhtuong wrote:
What is the value of $$A=\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+...+\frac{1}{48 \times 49 \times 50}$$?

A. $$\frac{306}{1225}$$

B. $$\frac{637}{2550}$$

C. $$\frac{1175}{4704}$$

D. $$\frac{1199}{4800}$$

E. $$\frac{2497}{9996}$$

In these kind of questions our strategy will always be to break down each term into difference of two terms, such that alternate terms cancel out. In thsi case we can do it as follows

$$A=\frac{1}{2} * \frac{2}{1 \times 2 \times 3}+\frac{2}{2 \times 3 \times 4}+...+\frac{2}{48 \times 49 \times 50}$$
$$A=\frac{1}{2} * \frac{3-1}{1 \times 2 \times 3}+\frac{4-2}{2 \times 3 \times 4}+...+\frac{50-48}{48 \times 49 \times 50}$$
$$A=\frac{1}{2} * ( \frac{3}{1 \times 2 \times 3} - \frac{1}{1 \times 2 \times 3} + \frac{4}{2 \times 3 \times 4} - \frac{2}{2 \times 3 \times 4}+...+\frac{50}{48 \times 49 \times 50} - \frac{48}{48 \times 49 \times 50} )$$
$$A=\frac{1}{2} * ( \frac{1}{1 \times 2 } - \frac{1}{2 \times 3} + \frac{1}{2 \times 3 } - \frac{1}{3 \times 4}+...+\frac{1}{48 \times 49} - \frac{1}{49 \times 50} )$$
Now alternate terms will cancel out and we will be left with only two terms
$$A=\frac{1}{2} * ( \frac{1}{1 \times 2} - \frac{1}{49 \times 50} )$$
$$A=\frac{1}{2} * ( \frac{1225-1}{49 \times 50} )$$
$$A=\frac{1}{2} * ( \frac{1224}{49 \times 50} )$$
$$A=\frac{306}{1225}$$

Hope it helps!
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Re: What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?  [#permalink]

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28 Nov 2016, 08:32
I think this is a 700 level question, its not easy to come up with a quick strategy where you can solve such a problem in almost 2.5 minutes
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What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?  [#permalink]

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Updated on: 16 Dec 2016, 13:10
For questions such as these with long multiplication or lots of arithmetic I prefer to use approximation.

A = 1/1*2*3 + 1/2*3*4 + 1/3*4*5 .. + .. 1/48*49*50
A = 1/6 + 1/24 + 1/60 ...
A = 5/24 + 1/60 ...

Now 5/24 > 4/24 > 1/6. Since the remaining numbers to be added are all relatively small I think it's fair to assume that our answer will be slightly larger than 1/6 but smaller than 1/4. So 1/6 < x < 1/4.

Out of the available answer choices the only one that comes close is 306/1225.

Originally posted by koenh on 14 Dec 2016, 07:43.
Last edited by koenh on 16 Dec 2016, 13:10, edited 1 time in total.
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Re: What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?  [#permalink]

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16 Dec 2016, 10:08
koenh wrote:
For questions such as these with long multiplication or lots of arithmetic I prefer to use approximation.

A = 1/1*2*3 + 1/2*3*4 + 1/3*4*5 .. + .. 1/48*49*50
A = 1/6 + 1/24 + 1/60 ...
A = 5/24 + 1/60 ...

Now 5/24 > 4/24 > 1/6. Since the remaining numbers to be added are all relatively small I think it's fair to assume that our answer will be slightly larger than 1/6 but smaller than 1/5. So 1/6 < x < 1/5.

Out of the available answer choices the only one that comes close is 306/1225.

But 5/24 is already > than 1/5, not smaller
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Re: What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?  [#permalink]

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21 Dec 2017, 14:35
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Re: What is the value of 1/(1*2*3)+1/(2*3*4)+...+1/(48*49*50)?   [#permalink] 21 Dec 2017, 14:35
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