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What is the value of 1+1/(1+2)+1/(1+2+3)

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What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 20 Feb 2018, 07:55
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What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(2\)
(B) \(\frac{100}{51}\)
(C) \(1\)
(D) \(\frac{50}{51}\)
(E) \(\frac{49}{50}\)


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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 20 Feb 2018, 09:06
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chetan2u wrote:
What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(\frac{49}{100}\)
(B) \(\frac{49}{50}\)
(C) \(1\)
(D) \(\frac{49}{25}\)
(E) \(\frac{49}{10}\)


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\(1= 2-\frac{2}{2}\)

\(\frac{1}{(1+2)}=\frac{2}{2}-\frac{2}{3}\)

\(\frac{1}{(1+2+3)}=\frac{2}{3}-\frac{2}{4}\) and so on, hence last term will be

\(\frac{1}{(1+2+3+....50)}=\frac{2}{50}-\frac{2}{51}\), on adding all the above numbers we will get

\(2-\frac{2}{51}=\frac{100}{51}\)

---------------------------------------------------------------

The formula for this type of series is \(\frac{2n}{(n+1)}\), where \(n\) is the number of term., here we have \(50\) terms so the addition should be

\(\frac{2*50}{(50+1)}=\frac{100}{51}\)
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What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post Updated on: 20 Feb 2018, 09:22
chetan2u wrote:
What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(\frac{49}{100}\)
(B) \(\frac{49}{50}\)
(C) \(1\)
(D) \(\frac{49}{25}\)
(E) \(\frac{49}{10}\)


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Hi Chetan2u--I am getting 100/51
if we will see the denominator : it is sum of consecutive integer :
\(\frac{n*(n+1)}{2}\)

the expression can be broken down into

\(\frac{2}{n*n(n+1)}\) -- where n = 1 to 50
further we can write it as\(\frac{2}{n}\) - \(\frac{2}{n+1}\)----- where n = 1 to 50

now 1st term\(\frac{2}{1}\) -\(\frac{2}{2}\)---1st
\(\frac{2}{2}\) - \(\frac{2}{3}\)--2nd
..
similarly it will go till
\(\frac{2}{50}\) - \(\frac{2}{51}\)---50th

each fraction wil be cancel out except 2 -\(\frac{2}{51}\) =\(\frac{100}{51}\)

Originally posted by sobby on 20 Feb 2018, 09:10.
Last edited by sobby on 20 Feb 2018, 09:22, edited 1 time in total.
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Re: What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 20 Feb 2018, 09:21
niks18 wrote:
chetan2u wrote:
What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(\frac{49}{100}\)
(B) \(\frac{49}{50}\)
(C) \(1\)
(D) \(\frac{49}{25}\)
(E) \(\frac{49}{10}\)


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\(1= 2-\frac{2}{2}\)

\(\frac{1}{(1+2)}=\frac{2}{2}-\frac{2}{3}\)

\(\frac{1}{(1+2+3)}=\frac{2}{3}-\frac{2}{4}\) and so on, hence last term will be

\(\frac{1}{(1+2+3+....50)}=\frac{2}{50}-\frac{2}{51}\), on adding all the above numbers we will get

\(2-\frac{2}{51}=\frac{100}{51}\)

Hi chetan2u,

can you clarify what I am missing here. also the formula for this type of series is \(\frac{2n}{(n+1)}\), where n is the number of term., here we have 50 terms so the addition should be

\(\frac{2*50}{(50+1)}=\frac{100}{51}\) same as my answer, if there had been only 1 term, then the sum would have been 2*1(1+1)=1, i.e. the first term in the series.



Hi..

you are absolutely correct..
Actually in hurry, I added the choices of a similar question I had made
Edited .. thanks and kudos
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 20 Feb 2018, 09:54
chetan2u wrote:
niks18 wrote:
chetan2u wrote:
What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(\frac{49}{100}\)
(B) \(\frac{49}{50}\)
(C) \(1\)
(D) \(\frac{49}{25}\)
(E) \(\frac{49}{10}\)


New question...
Kudos for best solutions


\(1= 2-\frac{2}{2}\)

\(\frac{1}{(1+2)}=\frac{2}{2}-\frac{2}{3}\)

\(\frac{1}{(1+2+3)}=\frac{2}{3}-\frac{2}{4}\) and so on, hence last term will be

\(\frac{1}{(1+2+3+....50)}=\frac{2}{50}-\frac{2}{51}\), on adding all the above numbers we will get

\(2-\frac{2}{51}=\frac{100}{51}\)

Hi chetan2u,

can you clarify what I am missing here. also the formula for this type of series is \(\frac{2n}{(n+1)}\), where n is the number of term., here we have 50 terms so the addition should be

\(\frac{2*50}{(50+1)}=\frac{100}{51}\) same as my answer, if there had been only 1 term, then the sum would have been 2*1(1+1)=1, i.e. the first term in the series.



Hi..

you are absolutely correct..
Actually in hurry, I added the choices of a similar question I had made
Edited .. thanks and kudos


Thanks chetan2u for updating.

Nonetheless its a great question :thumbup:
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What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 20 Feb 2018, 10:03
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chetan2u wrote:
What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(2\)
(B) \(\frac{100}{51}\)
(C) \(1\)
(D) \(\frac{50}{51}\)
(E) \(\frac{49}{50}\)


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We know that \(1+2+…+n\)=\(\frac{n(n+1)}{2}\)

Therefore, the nth term is = \(1/\frac{n(n+1)}{2}\)
or, nth term = \(2[\frac{1}{n} - \frac{1}{n+1}]\)

As you can see, when we add from n = 1 to n = 50, all numbers will cancel out accept -
\(2[1 - \frac{1}{51}] = \frac{100}{51}\)
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Re: What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 01 Mar 2018, 04:15
chetan2u, hello! Thank you for the question!
Accidentially I cannot understand the explanation, can you please explain the solution in details?
Thank you in advance!
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Re: What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 01 Mar 2018, 09:05
1
AkbarShakenov wrote:
chetan2u, hello! Thank you for the question!
Accidentially I cannot understand the explanation, can you please explain the solution in details?
Thank you in advance!


Hi AkbarShakenov

in this type of series question, you need to identify the patterns. Each term in the series will follow a similar pattern.

so first term of the series 1 can be written as 2-2/2

second term 1/(1+2)=1/3 can be written as 2/2-2/3. Note the relationship between first term and second term 2/2 is common in both the terms but has opposite signs. Hence when we will add all the individual terms, common terms will get cancelled. similarly you can re-write each term of the original series as a difference of two different numbers.

Depending upon the level of question, the pattern can be easy or difficult to identify. In most case the first two terms should be able to provide you a pattern.
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Re: What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 05 Mar 2018, 03:15
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niks18

Got it. Thank you!
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Re: What is the value of 1+1/(1+2)+1/(1+2+3)  [#permalink]

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New post 13 Mar 2018, 18:52
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I do not think the propose of this question is about calculation, If we see the equation carefully, it has 1 and other positive fractions in the equation, so we can let C,D,E out, because they are <=1, then we know that all the sums of the fractions can not be superior than 1, so let A out, we have only B to choose.
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Re: What is the value of 1+1/(1+2)+1/(1+2+3) &nbs [#permalink] 13 Mar 2018, 18:52
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