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

Question

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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niks18 · Accepted Answer

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}

gerry1881 · Answer

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.

sobby · Answer

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}

chetan2u · Answer

Hi..

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

niks18 · Answer

Thanks chetan2u for updating. Nonetheless its a great question

rahul16singh28 · Answer

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

As you can see, when we add from n = 1 to n = 50, all numbers will cancel out accept - 
 2  = \frac{100}{51}

AkbarShakenov · Answer

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!

niks18 · Answer

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.

AkbarShakenov · Answer

niks18

Got it. Thank you!

123456789007 · Answer

Therefore, the nth term is = 1/n(n+1)21/n(n+1)2
or, nth term = 2 2

Could you please explain me the simplification part of this step

fireagablast · Answer

Maybe not the best approach, but i recognized that it was essentially increasing powers of 2, but + 1.
1+(1/2^1)+(1/2^2) +(1/2^3)....

Since the +1 is negligible, i just threw it out.
Adding the first few terms, you quickly see that the numerator is "almost" double the denominator.
This leaves you with A or B, and B is the better option.

sanya2711 · Answer

You can use simple approximation over here. As it is evident that the value will be greater than 1, we can cut all options with values lesser than 1. The value also is less than 2 but greater than 1.6. Thus the correct answer is deducible in less than 30 seconds.

axbycz37 · Answer

I did it this way:

We know the following:
First term = 1
Second term = 1/3
Sum up to this point = 1 + 1/3 = 4/3
In all further terms, the denominator is increasing. As a result, the fraction tends closer to one as we move ahead in the series.

Now lets evaluate the options:
A - 2 (Seems too large as the fractions are diminishing with each and the initial fraction is less than half (1/3).
B - 100/51 - seems to fit in the criterion (Little under 2)
C - 1 - Which is the first term itself. Since we have no negative fractions, eliminate C.
D - Less than 1 - Eliminate
E - Less than 1 - Eliminate

So the only choice we are left with is B. Hence, I chose B.

shivangisharmaaa · Answer

why can they not be superior to 1??

chetan2u · Answer

I am sure, the member meant greater than 2 and not greater than 1.

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

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