sashibagra wrote:
iamheisenberg wrote:
Just want to post a general strategy to solve this kind of question. As this is already solved, I will try on another similar kind. Let's try to find out the value of the below expression:
\(\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}\)
The key to solve these type of questions is to rewrite each term as difference of two fractions and then cancel out maximum possible terms.
How can we rewrite each term as difference of two fractions??
1st term : \(\frac{1}{1*3} = (\frac{1}{2})\frac{2}{1*3} = (\frac{1}{2})\frac{3-1}{1*3} = (\frac{1}{2})(\frac{3}{1*3} - \frac{1}{1*3}) = (\frac{1}{2})(\frac{1}{1} - \frac{1}{3})\)
2nd term : \(\frac{1}{2*4} = (\frac{1}{2})\frac{2}{2*4} = (\frac{1}{2})\frac{4-2}{2*4} = (\frac{1}{2})(\frac{4}{2*4} - \frac{2}{2*4}) = (\frac{1}{2})(\frac{1}{2} - \frac{1}{4})\)
Similarly, 3rd term: \(\frac{1}{3*5} = (\frac{1}{2})(\frac{1}{3} - \frac{1}{5})\)
(n-1)th term: \(\frac{1}{(n-1)*(n+1)} = (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)})\)
nth term: \(\frac{1}{n*(n+2)} = (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})\)
Now using the above terms in the expression:
\(\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}\)
= \((\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) +(\frac{1}{2})(\frac{1}{4} - \frac{1}{6}) + ........ + (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)}) + (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})\)
= \((\frac{1}{2}) (\frac{1}{1}\) \(- \frac{1}{3}\) \(+ \frac{1}{2}\) \(- \frac{1}{4}\) \(+ \frac{1}{3}\) \(- \frac{1}{5}\) + \(\frac{1}{4}\) \(- \frac{1}{6} + ....... + \frac{1}{(n-2)}\) \(- \frac{1}{n}\) + \(\frac{1}{(n-1)} - \frac{1}{(n+1)}\) \(+ \frac{1}{n}\) \(- \frac{1}{(n+2)})\)
In the above expression, you can notice that the same coloured terms can be cancelled out.
Now after cancelling we can simplify the complete expression to
\((\frac{1}{2}) (\frac{1}{1} + \frac{1}{2} - \frac{1}{(n+1)} - \frac{1}{(n+2)})\)
We can use similar concept to such kind of problems.
Is this helpful?
I have doubt about it!
Ain't it time consuming and how much time did you consumed to solve this math your way? ( Just curious so would be better if you mention time needed to solve your way of thinking. I appreaciate thinking about possibilities but if during GMAT exam we are under time constrain so students always go after those techniques that helps them to pick the right answer in best possible time.
Thanks.
Not at all. It didn't take much time - probably a minute or less. It's just the way I explained it in the question is lengthy. Otherwise, once you get this concept, I think it is much easier, quicker and more accurate to do this way than the alternative way of calculating each term's approx value and summing it up.
Actually I had solved similar problems many years back in school and I knew the concept. So as soon as I saw this question I knew how to approach. May be that's why it took only a minute or less for me. And that is the reason I wanted to share the approach to such kind of questions (not only to this question).
But there are always different methods to solve a problem. And other people may find it easier to calculate the value of each term, sum them and check the answer choices.