Archive - Dec 11, 2007

Date

Don't do proofs on time limited math contests

Posted December 11th, 2007 by Mgccl
in

I was encountered with a question in the matheletes, a small math contest organized in our county every few weeks. We have 3 pair of questions, 12 minute for each pair. One of the questions as follows.
Simplify the equation below
\frac{1}{1 \times 2}+ \frac{1}{2 \times 3}+ \frac{1}{3 \times 4}{...}+ \frac{1}{99\times100}
Anyone who have the slightest idea of doing contests in a limited amount of time knows prove a theorem is not the way to solve questions. test first few numbers and the answer would be clear, 99/100.
I was so stupid and trying to solve it by generalize the equation.
The equation simplified in summation form
\sum _{k=0}^{98} \frac{1}{(1+k)(2+k)}
Then, change 98 into n
\sum _{k=0}^{n} \frac{1}{(1+k)(2+k)}
First, condense the equation into one fraction form
\frac{\sum_{k=0}^{n}\frac{\prod_{i=0}^{n}(1+i)(2+i)}{(1+k)(2+k)}} {\prod_{i=0}^{n}(1+i)(2+i) }
Solve for the products.(which is not necessary )
\frac{\sum_{k=0}^{n}\frac{(1+n)!(2+n)!}{(1+k)(2+k)}} {(1+n)!(2+n)!}
that's when I found that the products are constant, so I took the result of the product outside the summation.
\frac{(1+n)!(2+n)!\sum_{k=0}^{n}\frac{1}{(1+k)(2+k)}} {(1+n)!(2+n)!}
now, cross out the common denominator and I'm back with the original equation.
That useless proof took way too much time(time to formulate and write on paper), so I did not have the chance to just test the first few numbers and recognize the pattern.
I learned a important lesson, where there is a time limit, don't try to do proofs.
I'm still working on the proof, (because I can't find it anywhere else), I have to try a different approach. If anyone have suggestions or know how to prove it, please leave a comment, thx.

Update: I have got the answer on the mathlinks forum

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