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I recently came across this question, and I need some help with it

$(1- 1/2)\times(1-1/3)\times(1-1/4) ... (1-1/2016)\times (1-1/2017)=$

I have worked out that the pattern goes :

$1\over2$$\times$$2\over3$$\times$$3\over4$...

The last fraction will be $2016\over2017$, as with every fraction, the numerator and denominator increase by 1. However, how do I manage to multiply these?

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  • $\begingroup$ Just multiply out a few more fractions to check the surviving terms $\endgroup$ Commented May 17, 2017 at 9:39
  • $\begingroup$ Is this yet another attempt to cheat in the Australian Junior Contest? Like here? $\endgroup$ Commented May 20, 2017 at 7:43

3 Answers 3

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The product telescopes: the terms cancel, like so: $$\frac{1}{\not 2}\times\frac{\not 2}{\not 3}\times \dots\times \frac{\not{2016}}{2017}.$$ Thus the answer is $\frac{1}{2017}$.

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Hint: We know that $\frac{a}{b}\times\frac{c}{d}=\frac{ac}{bd}=a\times\frac{c}{b}\times\frac{1}{d}$. Try using that fact and see if any terms can cancel out.

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$$\prod_{n=2}^{2017}\left(1-\frac1n\right)=\prod_{n=2}^{2017}\frac{n-1}{n}=\frac{\prod_{n=2}^{2017}(n-1)}{\prod_{n=2}^{2017}n}= \frac{\prod_{n=1}^{2016}n}{\prod_{n=2}^{2017}n}=\frac{1\cdot \prod_{n=2}^{2016}n}{2017\cdot \prod_{n=2}^{2016}n}=\frac1{2017}$$

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