Method-1

\(\frac{1}{(1*2)} + \frac{1}{(2*3)} + \frac{1}{(3*4)} + \frac{1}{(4*5)} + \frac{1}{(5*6)} + \frac{1}{(6*7)} + \frac{1}{(7*8)} + \frac{1}{(8*9)} + \frac{1}{(9*10)}\)

Can be Rewritten as

\([\frac{1}{1}-\frac{1}{2}] + [\frac{1}{2}-\frac{1}{3}] + [\frac{1}{3}-\frac{1}{4}] + [\frac{1}{4}-\frac{1}{5}] + [\frac{1}{5}-\frac{1}{6}] + [\frac{1}{6}-\frac{1}{7}] + [\frac{1}{7}-\frac{1}{8}] + [\frac{1}{8}-\frac{1}{9}] + [\frac{1}{9}-\frac{1}{10}]\)

Every terms cancels out except

\([\frac{1}{1}-\frac{1}{10}] = \frac{9}{10}\)
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Tn = nth term of the series = 1/(n)(n+1) = (n+1-n)/(n)(n+1) = 1/n - 1/(n+1)

If you add all the terms of this series (to n terms), you get a sum of 1 - 1/(n+1) as all the other terms cancel out.
So sum of 9 terms = 1 - 1/10 = 9/10.
Option (E).
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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?

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.

RIght off the bat, you can see that 1/2 + 1/4 = 3/4. So, the answer is already greater that choices A), B) or C). Based on the acscending nature of the answer choices, we can conclude that 46/55 is less than 9/10. But just using some reasoning, we are already at .75, 1/90 is a bit more than .01, 1/30 is a bit more than .03, 1/20 is .05. Adding these into .75, we are already at .84. Knowing the other 3 terms will give us approximately another .05, we can safely choose answer choice E.
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Sum = mean * number of terms

The mean (1/5*6) quoted by you is incorrect.



Posted from my mobile device
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