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

AI Thread Summary
The discussion revolves around solving the mathematical expression (1−1/2)(1−1/3)...(1−1/2017) and finding its value, x, suitable for Year 7 and 8 students. Participants share their solutions and express confusion about the complexity of the problem relative to their educational experience. There is a request for clarification on the axioms, theorems, and definitions applicable to the proof used in solving the expression. The conversation highlights the perceived difficulty of the problem for the intended age group. Overall, the thread emphasizes the need for accessible explanations of mathematical concepts for younger students.
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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
What are the axiom(s) or theorem(s) and definition(s) that are used in the above proof that one learns in years 7 and 8 in mathematics ??
 
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