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.
bio
Messages
12
Reaction score
0
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!
 
Last edited:
Mathematics news on Phys.org
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
 
Last edited by a moderator:
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 ??
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top