MHB Year 7 & 8 mathematics puzzle solve (1−1/2)(1−1/3)....(1−1/2017)=x

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This will be fairly easy to do, even in the eyes of a Year 7

Find x when $$\left({1-\frac{1}{2}}\right)\left({1-\frac{1}{3}}\right)...\left({1-\frac{1}{2017}}\right)=x$$

Remember: Year 7 and 8 maths only!
 
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My solution (to the problem originally posted):

I don't know where $x$ fits into this, however, by inspection, I find:

$$S=\sum_{k=1}^{2016}\left(\frac{k}{k+1}\right)=2017-H_{2017}$$

Where $H_n$ is the $n$th harmonic number.

According to W|A, the actual value of the given sum is:

$$S=\frac{822115465324282561724802258557532366310699004754387291526159866017130476270647843871743685992495701228849000100287286632619820316667316718584369564259306482609128747444592930143458010597849621845241330694119277917820930378101615543893613935369526817385751154419464839762367230184496904994441405722157454834997752398782223977479947678657020025523575193386812429482328568636994034136498298967034748318886404333894953886987231864341318926681248859492981091379567361347779648984940411352335500719085073786842132187244436822275817441998400895963445964438102195433232177881021740273056977601662086595538286853043320690576343454001212039307920418455812365948698036852323924962865025990303576841043964865867962487830471679713735545302265107731707712379650082724108349916371588495856107821003528476225549143904540772209314928517912601766945839363224442605754719911769816595440935014794609}{409254318735421372874527513713809849729135059223982154744686910391244881751394287792064717538513501909093551073331105901230174815690870254022597927796335026690598911658288578341662007257908728685152761650582748164977706081859581162971618587292791162956045102947117094844113969603097500003652004069652979380609518121519013531548885927853149410724755210783428395488365803795111980933624569399933400969497574721790682373349842950967442946949750794524393431836515674176033082971320894446597931131516693687546474012255765758283940350898733234638365606173236127739051742487602390618454018049479926800452950875570239516276819871570674576851400791100528044895551810310590444123249637327298475483510172556378112704981184569797419716363712359628331775989191307093772005846389088398901747521145065831525586976740612039273283797156150977660434534795320274675464655132985044254833228160000}$$

The decimal approximation is:

$$S\approx2008.813169924717623749873055650937838644716733197521974209$$
 
Re: Year 7 & 8 mathematics puzzle

This is my solution:
We can deduce that the pattern goes like $$\frac{1}{2},\frac{2}{3}...\frac{2016}{2017}$$ so, by cancelling, we get:$$\frac{1}{\cancel 2} \cdot \frac{\cancel 2}{\cancel 3} \cdot ... \frac{\cancel{2016}}{2017}$$

So the only remaining fraction is:

$$\frac{1}{2017}$$

- - - Updated - - -

Also, sorry about the coding problem. I can't manage to only make the code into italics. If anyone can edit it, please do
 
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This makes me wonder if my school ripped me off... I do not recall doing this sort of thing at year 7 and 8 level xD. No wonder I'm such a dunce :p.
 
If anyone can help me with my other questions, please do. Thanks ;)
 
Re: Year 7 & 8 mathematics puzzle

bio said:
This is my solution:
We can deduce that the pattern goes like $$\frac{1}{2},\frac{2}{3}...\frac{2016}{2017}$$ so, by cancelling, we get:$$\frac{1}{\cancel 2} \cdot \frac{\cancel 2}{\cancel 3} \cdot ... \frac{\cancel{2016}}{2017}$$

So the only remaining fraction is:

$$\frac{1}{2017}$$
Also, sorry about the coding problem. I can't manage to only make the code into italics. If anyone can edit it, please do
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