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?