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

[bio]

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!

 

[MarkFL]

My solution (to the problem originally posted):

Spoiler

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

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

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

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

\(\displaystyle S=\frac{822115465324282561724802258557532366310699004754387291526159866017130476270647843871743685992495701228849000100287286632619820316667316718584369564259306482609128747444592930143458010597849621845241330694119277917820930378101615543893613935369526817385751154419464839762367230184496904994441405722157454834997752398782223977479947678657020025523575193386812429482328568636994034136498298967034748318886404333894953886987231864341318926681248859492981091379567361347779648984940411352335500719085073786842132187244436822275817441998400895963445964438102195433232177881021740273056977601662086595538286853043320690576343454001212039307920418455812365948698036852323924962865025990303576841043964865867962487830471679713735545302265107731707712379650082724108349916371588495856107821003528476225549143904540772209314928517912601766945839363224442605754719911769816595440935014794609}{409254318735421372874527513713809849729135059223982154744686910391244881751394287792064717538513501909093551073331105901230174815690870254022597927796335026690598911658288578341662007257908728685152761650582748164977706081859581162971618587292791162956045102947117094844113969603097500003652004069652979380609518121519013531548885927853149410724755210783428395488365803795111980933624569399933400969497574721790682373349842950967442946949750794524393431836515674176033082971320894446597931131516693687546474012255765758283940350898733234638365606173236127739051742487602390618454018049479926800452950875570239516276819871570674576851400791100528044895551810310590444123249637327298475483510172556378112704981184569797419716363712359628331775989191307093772005846389088398901747521145065831525586976740612039273283797156150977660434534795320274675464655132985044254833228160000}\)

The decimal approximation is:

\(\displaystyle S\approx2008.813169924717623749873055650937838644716733197521974209\)

 

[bio]

Re: Year 7 & 8 mathematics puzzle

This is my solution:

Spoiler

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

 

[Joppy]

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.

 

[bio]

If anyone can help me with my other questions, please do. Thanks ;)

 

[solakis1]

Re: Year 7 & 8 mathematics puzzle

This is my solution:

Spoiler

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 ??