MHB Can You Solve This High School Math Series Without a Calculator?

  • Thread starter Thread starter anemone
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AI Thread Summary
The discussion announces a temporary hiatus from posting, with results for High School POTW #351 set to be released on February 12th. In the meantime, POTW #352 is introduced, challenging members to evaluate a complex mathematical expression without a calculator. Participants are encouraged to refer to the guidelines for submitting solutions. Congratulations are extended to members who successfully solved the previous problem. The thread emphasizes engagement with the math community through these weekly challenges.
anemone
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Hi to all members of MHB!

I won't be posting anything starting tomorrow onwards until February 12th, therefore, the result for High School POTW #351 will only be released on February 12th and I will post the POTW#352 (which is supposed to be posted next Tuesday@ February 5th) today too.

I hope you will enjoy solving these two weeks of high school problems I posted in a row.(Blush)Here is POTW #352:

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Without using a calculator, evaluate $$\frac{(4\times 7+2)(6\times 9+2)(8\times 11+2)\cdots(2016\times 2019+2)}{(5\times 8+2)(7\times 10+2)(9\times 12+2)\cdots(2015\times 2018+2)}$$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. kaliprasad
3. lfdahl

Solution from lfdahl:
We have,

\[ \frac{(4\cdot 7+2)(6\cdot 9+2)(8\cdot 11+2)\cdot ...\cdot (2016\cdot 2019+2)}{(5\cdot 8+2)(7\cdot10+2 )(9\cdot 12+2)\cdot ...\cdot (2015\cdot 2018+2)} \\\\=(2017\cdot 2020+2)\prod_{k=2}^{1008}\frac{2k(2k+3)+2}{(2k+1)(2k+4)+2} \\\\=((2018-1)\cdot (2019+1)+2)\prod_{k=2}^{1008}\frac{k(2k+3)+1}{(2k+1)(k+2)+1} \\\\=(2018\cdot 2019+2018-2019-1+2)\prod_{k=2}^{1008}\frac{(2k+1)(k+1)}{(2k+3)(k+1)}\\\\= 2018\cdot2019 \cdot \prod_{k=2}^{1008}\frac{2k+1}{2k+3} \\\\= 2018 \cdot 2019\cdot \frac{5}{7}\cdot \frac{7}{9}\cdot\frac{9}{11}\cdot ...\cdot \frac{2015}{2017}\cdot \frac{2017}{2019} \\\\=2018\cdot 2019\cdot \frac{5}{2019} \\\\= 10090\]
 
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