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    (Original post by justinawe)
    Seems like a fairly simple induction proof, what exactly are you having trouble with?
    What I don't get is what is all that '2n+1' and '2n' about when trying to prove something is even or odd?
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    (Original post by priyashah95)
    So scared for tomorrow wanna cry


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    dont worry
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    (Original post by ThePersian)
    What I don't get is what is all that '2n+1' and '2n' about when trying to prove something is even or odd?
    Oh, I see.

    2n means you have an integer n multiplied by two... which means it is divisible by two. By definition, an even number has to be divisible by two, and therefore an even number can always be represented by "2n".

    Now, we know that an even number+1=an odd number, so an odd number can be represented by 2n+1.
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    (Original post by ThePersian)
    Hello,

    Could someone help me out here, I never understand proof questions.
    Attachment 215104

    If you know, can you also explain this whole idea behind '2n+1' and '2n' I don't understand it.

    Thank you.
    You don't need to write it in the form of 2n+1/ 2n/ whatever.

    n^{3} + 3n^{2} + 2n = n(n+1)(n+2), n \in \mathbb{Z^{+}}

    As 'n' is a positive integer, the expression is the product of 3 consecutive integers - so you have factors of 2 and 3 in there, thus it is divisible by 6 for all n.
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    (Original post by Felix Felicis)
    You don't need to write it in the form of 2n+1/ 2n/ whatever.

    n^{3} + 3n^{2} + 2n = n(n+1)(n+2), n \in \mathbb{Z^{+}}

    As 'n' is a positive integer, the expression is the product of 3 consecutive integers - so you have factors of 2 and 3 in there, thus it is divisible by 6 for all n.
    I think he was asking about 2n/2n+1 thing was a side question rather than as part of that question
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    (Original post by justinawe)
    I think he was asking about 2n/2n+1 thing was a side question rather than as part of that question
    Oh! I see, my bad
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    Is it just me or is no one else in the mood to revise for m1 tomorrow?
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    (Original post by Felix Felicis)
    You don't need to write it in the form of 2n+1/ 2n/ whatever.

    n^{3} + 3n^{2} + 2n = n(n+1)(n+2), n \in \mathbb{Z^{+}}

    As 'n' is a positive integer, the expression is the product of 3 consecutive integers - so you have factors of 2 and 3 in there, thus it is divisible by 6 for all n.
    Thank you!
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    (Original post by DJMayes)
    Double quoted!

    Firstly, Edexcel are notoriously bad at exam timetabling, for no conceivable reason other than to finish AS Level exams before A2 level exams. For example, I have the following clashes in my Maths exams this summer:

    18th June - C4/M4
    21st June - FP2/S4
    24th June - FP3/M5

    So many clashes could be avoided just by, for example, sitting M5 with M1 instead of with FP3, as M5 and FP3 are basically always sat together. But Edexcel never do this.

    And you'll find calculus comes up in all modules past M1. M2 has basic differentiation and integration to solve kinematics problems. M3 contains more advanced differentiation and integration in kinematics, including acceleration as a function of displacement, integration to find work done and impulse, and integration to find centres of mass. M4 contains first and second order differential equations, along with calculus to find positions of stability. M5 contains first and second order vector differential equations, differential equations on varying mass (Where you have to derive the equations yourself using limits), integration to find moments of inertia and a bit of calculus in rotational motion.
    Thanks for the reply. It was very informative! Can I post the post in the M1 thread the post you made a few days ago on M1 in this thread for the benefit of M1 candidates who don't follow this thread?
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    (Original post by reubenkinara)
    Thanks for the reply. It was very informative! Can I post the post you made a few days ago on M1 in the M1 thread for the benefit of M1 candidates who don't follow this thread?
    I haven't posted in the M1 thread... :confused:

    EDIT: This has now been explained to me by the mysterious forces watching over this thread. Yeah, go ahead, the post is meant to be helping people. I couldn't really care less if someone posted it there claiming it as their own to be quite honest, so long as someone gets something out of it.
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    (Original post by Ursin)
    Yo, I wonder if anyone can answer this S1 question? I can't do it, my teachers can't do it, my friend who got 100 in S1 the first time we took it can't.
    Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
    a) How many different sequences of 8 cards are possible?
    For this I did 52^8, which is correct.
    b) How many of the sequences in part a) will contain 3 picture cards, 3 odd-numbered cards and 2 even-numbered cards?
    The answer in the back of the book says 3.097 x 10^12. I have tried a multitude of methods and cannot for the life of me obtain this answer.
    Thanks
    I can't get that answer either. Are you sure it is correct?

    Perhaps the man himself may shed a light on this.

    (Original post by ghostwalker)
    ...
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    (Original post by Ateo)
    I can't get that answer either. Are you sure it is correct?

    Perhaps the man himself may shed a light on this.
    (Original post by Ursin)
    Yo, I wonder if anyone can answer this S1 question?...
    Ok, call the sequence of picture cards, odd cards and even cards: PPPOOOEE
    As there are 12 picture cards, 20 odd cards, 20 even cards, then the number of ways this specific sequence can be made is:

    12^{3} \times 20^{3} \times 20^{2}

    But this is simply one of the ways we can arrange this sequence; we could have POPOPOEE for example as well, this still counts as 3 pictures, 3 odds, 2 evens. There are 8 items, so the total number of ways we can arrange this sequence is:

    12^{3} \times 20^{3} \times 20^{2} \times 8!

    But we've over-counted here - for example, the sequence of say PPP and PPP are the exact same. Since there are 3 Ps, 3 Os, 2 Es, we've over counted by 3! \times 3! \times 2!

    Hence, the total number of ways this sequence can be made is \dfrac{12^{3} \times 20^{3} \times 20^{2} \times 8!}{3! \times 3! \times 2!} = 3.096576 \times 10^{12}
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    Hey guys, I've got two questions on M1.

    1) When the question asks you to "write an equation for motion" of something, what's the best way to go about this? Do they care about how you write it? i.e. Resolving a particle of mass 4m hanging off of a string in the downards direction in equilibrium, including tension, could it be 4m - T = 0 or does it have to be 4m = T?

    2) Can you use Fmax = uR? This was the way I was taught friction in school, but as soon as I saw DJ's post, this confused me a bit.

    Thanks!
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    (Original post by Bixel)
    Hey guys, I've got two questions on M1.

    1) When the question asks you to "write an equation for motion" of something, what's the best way to go about this? Do they care about how you write it? i.e. Resolving a particle of mass 4m hanging off of a string in the downards direction in equilibrium, including tension, could it be 4m - T = 0 or does it have to be 4m = T?

    2) Can you use Fmax = uR? This was the way I was taught friction in school, but as soon as I saw DJ's post, this confused me a bit.

    Thanks!
    If something is in motion then \ F_{Max} = \mu R
    I think either's satisfactory for your first q, though I don't actually know.
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    Y'all recon we will get a long question on connected particles?
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    (Original post by Bixel)
    Hey guys, I've got two questions on M1.

    1) When the question asks you to "write an equation for motion" of something, what's the best way to go about this? Do they care about how you write it? i.e. Resolving a particle of mass 4m hanging off of a string in the downards direction in equilibrium, including tension, could it be 4m - T = 0 or does it have to be 4m = T?

    2) Can you use Fmax = uR? This was the way I was taught friction in school, but as soon as I saw DJ's post, this confused me a bit.

    Thanks!
    1) I shouldn't have thought it matters, but someone else might be able to shed light - I think so long as the content is right, it should be fine, unless they've asked for a certain form!
    2) Fmax=uR when a particle (or whatever) is moving or on the point of moving (limiting equilibrium). In M1 the majority of questions do say that, but if the object/particle is stationary and the question doesn't say it's on the point of moving then that isn't the maximum value of Friction necessarily, which is where the inequality Fr<=uR comes into play.
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    (Original post by Felix Felicis)
    ...
    What puzzles me is why does the answer of the book for the first part include various arrangements of the same sequence and we take into account the fact that the same sequence could be arranged in different ways in the second part?
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    (Original post by joostan)
    If something is in motion then \ F_{Max} = \mu R
    I think either's satisfactory for your first q, though I don't actually know.
    Hmm, I think I get it now. Would I be right to say that, when  \[F &lt; \mu R\] , the object does not move, yet when F \geqslant  \mu R, the object moves?
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    (Original post by Bixel)
    Hmm, I think I get it now. Would I be right to say that, when  \[F &lt; \mu R\] , the object does not move, yet when F \geqslant  \mu R, the object moves?
    Your first point is correct. However,  F= \mu R does not necessarily imply that the object is in motion - it could be in limiting equilibrium.
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    (Original post by L'Evil Fish)
    Yeah there are more... But didn't want to list everyone :lol:

    Who's ready for their C1/M1 tomorrow?
    Ahh this takes me back, no better way to start Maths A Level than with C1 and M1 :')
 
 
 
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