What is the sum of 1/1∗4 + 1/2∗5 + 1/3∗6 + 1/4∗7.... ?

Question

What is the sum of  \frac{1}{1∗4} + \frac{1}{2∗5} + \frac{1}{3∗6} + \frac{1}{4∗7}  +... ?

A. 9/17
B. 7/15
C. 11/18
D. 92/173
E. 15/82

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chetan2u · Answer

Try to find a pattern.

Each of the term is of type  \frac{1}{n(n+3)} , which can be written as  \frac{1}{3}*(\frac{1}{n}-\frac{1}{n+3})

\frac{1}{1∗4} + \frac{1}{2∗5} + \frac{1}{3∗6} + \frac{1}{4∗7}  +... 
 \frac{1}{3}(\frac{1}{1}-\frac{1}{4}) + \frac{1}{3}(\frac{1}{2}-\frac{1}{5}) +\frac{1}{3}(\frac{1}{3}-\frac{1}{6}) + \frac{1}{3}(\frac{1}{4}-\frac{1}{7}) + \frac{1}{3}(\frac{1}{5}-\frac{1}{8})

\frac{1}{3}(\frac{1}{1}-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\frac{1}{5}-\frac{1}{8})+….

\frac{1}{3}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+………-(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+….)

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

\frac{1}{3}(\frac{11}{6})=\frac{11}{18}

C

Andrew89 · Answer

Is it possible to counter such a question in real GMAT ? it seems impossible to solve it in 2 mins

Rohit_842 · Answer

How does one even approach such problems, they always knock me out. Please help  HarshavardhanR

HarshavardhanR · Answer

Hey  Rohit_842 ,

This is a somewhat tricky question (I think!). The key is to find a pattern that will help simplify what is given.

Key Conceptual Idea  ->

Typically, a term which looks like  \frac{<Numerator>}{(a)*(b)}  can be converted into a combination of multiple terms where the denominator contains only "a" or only "b".

For instance ->

\frac{1}{2*3}  = 1/2 - 1/3

\frac{1}{4*10}  = (1/4 - 1/10) * (1/6)

Any such term can be converted into a combination of terms that avoid multiple numbers in the denominator.

Why does this help?  Because, in a lot of cases, dealing with these simpler terms is easier (in a lot of these questions, many of these terms cancel each other out). Hence, this is one of the ways to simplify terms like the ones given in our question.  Remember this!

Now, let us apply the above conceptual idea to solve this question.

(1) Observe the given sum

\frac{1}{1*4}  +  \frac{1}{2*5}  +  \frac{1}{3*6}  + .....

Observe every term.

We can see that every term is of the format ->   \frac{1}{(n)(n+3)} 
 
 (2) Finding a pattern that can help simplify the given sum

\frac{1}{(n)(n+3)}  = ( \frac{1}{n}  -  \frac{1}{(n+3)} ) *  \frac{1}{3}

(3) Using the above pattern to simplify the sum

\frac{1}{1*4}  +  \frac{1}{2*5}  +  \frac{1}{3*6}  + .....

=  \frac{1}{3}  * (1 - 1/4 + 1/2 - 1/5 + 1/3 - 1/6 + 1/4 - 1/7 + 1/5 - 1/8 + 1/6 - 1/9 + ......)

=  \frac{1}{3}  *  (1/4 + 1/5 + 1/6 + ....)  -  (1/4 + 1/5 + 1/6 + .....)  ]

Terms in bold cancel each other!

=  \frac{1}{3}  * (1 +  \frac{1}{2}  +  \frac{1}{3} ) 
=  \frac{1}{3}  *  \frac{11}{6}  =  \frac{11}{18} .  Choice C.

Hope this helps!
Harsha

Kaifbinarshad · Answer

this is how i solved it,

1/4+1/10+1/18+1/28+.....

we can rewrite it in decimal form,
0.25+0.1+approx 0.2+ approx. 0.3,
which gives us approx. 0.8 something only option A, C and D are above 50% and out of them option C is the highest and thus our answer.

What is the sum of 1/1∗4 + 1/2∗5 + 1/3∗6 + 1/4∗7.... ?

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GMAT Problem Solving (PS) Questions

What is the sum of 1/1∗4 + 1/2∗5 + 1/3∗6 + 1/4∗7.... ?

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