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

[anemone]

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!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. kaliprasad
3. lfdahl

Solution from lfdahl:

Spoiler

We have,

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