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    (Original post by joostan)
    I'm sorry to say that the two forms you've given me are not equivalent. Check the bold.
    The inductive proof was so much tidier
    (Original post by DJMayes)
    As said, it cost me 5 UMS - I don't know exactly how many raw marks that is though, and I'm fairly sure I got the rest of the paper right.



    You're not allowed to just quote the summation formula, so there are a few things you could do. Firstly, you could prove the summation formula  S_n = \frac{1}{2}n(2a+(n-1)d) (I think that's the formula) but in my opinion a better way would be to simply prove the given summation using induction.

    (You could also prove the result for the sum from one to n of r, and then use this and series methods to derive the result. The problem is that I'm not familiar with the WJEC syllabus so I don't know what method is going to be expected.)
    Wait what?

    I can get the summation from first principles...

    But then what do I do with that?

    ( 4n^2 - 2n ) / 2 = 2n^2 - n

    = n(2n-1)
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    (Original post by L'Evil Fish)
    Wait what?

    I can get the summation from first principles...

    But then what do I do with that?

    ( 4n^2 - 2n ) / 2 = 2n^2 - n

    = n(2n-1)
    You ask a question, then proceed to solve it... so are you really asking a question at all? :curious:
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    (Original post by justinawe)
    You ask a question, then proceed to solve it... so are you really asking a question at all? :curious:
    :dontknow:

    So my two choices are:

    Prove general summation and sub in values.

    Prove summation using the values there (as Llewellyn showed me from the mark scheme)
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    (Original post by L'Evil Fish)
    :dontknow:

    So my two choices are:

    Prove general summation and sub in values.

    Prove summation using the values there (as Llewellyn showed me from the mark scheme)
    I honestly have no clue what is meant by proving "from first principles" here, so I'd just go with what they put in the markscheme.
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    (Original post by justinawe)
    I honestly have no clue what is meant by proving "from first principles" here, so I'd just go with what they put in the markscheme.
    Okay that's fine then...

    What else... There was this too, maybe I'm just finding manipulation difficult, it was last night though...

    I'll post them in a few minutes if you don't mind?
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    (Original post by L'Evil Fish)
    Okay that's fine then...

    What else... There was this too, maybe I'm just finding manipulation difficult, it was last night though...

    I'll post them in a few minutes if you don't mind?
    Sure, go ahead.
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    (Original post by justinawe)
    Sure, go ahead.
    Really simple method I guess... Just don't know where I went wrong.

    AP:
    55 = Sum of first ten
    4th + 7th + 9th = 27

    55 = 5(a+9d)
    11 = a + 9d
    a = 11 - 9d

    (a+3d)+(a+6d)+(a+8d)=27
    (11-9d+3d)+(11-9d+6d)+(11-9d+8d)=27

    33 - 10d = 27
    d = 6/10
    a = 5.6
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    (Original post by joostan)
    On second thoughts I suppose you're right.
    Btw I proved the summation formula on the previous page
    Ahh, fair do's, fair do's ^^
    (Original post by L'Evil Fish)
    :dontknow:

    So my two choices are:

    Prove general summation and sub in values.

    Prove summation using the values there (as Llewellyn showed me from the mark scheme)
    Or, choice number 3... :emo:
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    (Original post by L'Evil Fish)
    Really simple method I guess... Just don't know where I went wrong.

    AP:
    55 = Sum of first ten
    4th + 7th + 9th = 27

    55 = 5(a+9d)
    11 = a + 9d
    a = 11 - 9d

    (a+3d)+(a+6d)+(a+8d)=27
    (11-9d+3d)+(11-9d+6d)+(11-9d+8d)=27

    33 - 10d = 27
    d = 6/10
    a = 5.6
    2a
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    (Original post by justinawe)
    2a
    Damn, okay, that's sorted then...

    Last one before it should be full marks in my knowledge (like I could now)

    I think it was a circle one...

    Area of Sector OAB = 43.56

    Arc BC is 13cm shorter than CD.

    Radius is 11cm.

    I need to find theta and the other one.

    I got

    Theta = 0.72
    The other one = 1.80 (this is wrong I think)

    Forgot to post a diagram... *walks upstairs*

    And now it wont upload :mad:
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    (Original post by L'Evil Fish)
    Damn, okay, that's sorted then...

    Last one before it should be full marks in my knowledge (like I could now)

    I think it was a circle one...

    Area of Sector OAB = 43.56

    Arc BC is 13cm shorter than CD.

    Radius is 11cm.

    I need to find theta and the other one.

    I got

    Theta = 0.72
    The other one = 1.80 (this is wrong I think)
    Your value for theta is right.

    What's this "other one" you speak of, though?
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    (Original post by justinawe)
    Your value for theta is right.

    What's this "other one" you speak of, though?
    Call it angle x.

    Basically:

    11x + 13 = 11(pi-x)

    In my opinion, because it's the diameter, split up into two sectors, with one labelled x. And one arc is 13cm less than the other arc.
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    (Original post by L'Evil Fish)
    Wait what?

    I can get the summation from first principles...

    But then what do I do with that?

    ( 4n^2 - 2n ) / 2 = 2n^2 - n

    = n(2n-1)
    Precisely
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    (Original post by joostan)
    Precisely
    Sn = 1 + 5 + . . . + (4n – 7) + (4n – 3)
    Sn = (4n – 3) + (4n – 7) + . . . + 5 + 1
    Reversing and adding
    Either:
    2Sn = (4n – 2) + (4n – 2) + . . . + (4n – 2) + (4n – 2)
    Or:
    2Sn = (4n – 2) + . . . (n times)
    2Sn = n(4n – 2)
    Sn = n(2n – 1)
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    (Original post by L'Evil Fish)
    Sn = 1 + 5 + . . . + (4n – 7) + (4n – 3)
    Sn = (4n – 3) + (4n – 7) + . . . + 5 + 1
    Reversing and adding
    Either:
    2Sn = (4n – 2) + (4n – 2) + . . . + (4n – 2) + (4n – 2)
    Or:
    2Sn = (4n – 2) + . . . (n times)
    2Sn = n(4n – 2)
    Sn = n(2n – 1)
    Yeah or . . .
    Option 3:
    Spoiler:
    Show
    \ u_n = 4n -3 Show that \ S_n = n(2n-1)
    Basis Case for S_n
     n = 1 \Rightarrow S_1 = u_1 = 1 = 1(2\times 1 - 1)
    Assume true for:
    n = k \Rightarrow S_k  = k(2k-1) = 2k^2 -k
    Now consider:
     n = k+1 \Rightarrow S_{k+1} = (k+1)(2(k+1) - 1) = 2k^2 +3k +1
    We also know that:
    \ S_{k+1} = S_k + u_{k+1}
    As
    u_{k+1} = 4(k+1) - 3 = 4k +1

 \Rightarrow S_{k+1} = 2k^2 - 1 + (4k+1) = 2k^2 + 3k +1
    Blah blah blah proved by induction etc.
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    (Original post by joostan)
    Yeah or . . .
    Option 3:
    Spoiler:
    Show
    \ u_n = 4n -3 Show that \ S_n = n(2n-1)
    Basis Case for S_n
     n = 1 \Rightarrow S_1 = u_1 = 1 = 1(2\times 1 - 1)
    Assume true for:
    n = k \Rightarrow S_k  = k(2k-1) = 2k^2 -k
    Now consider:
     n = k+1 \Rightarrow S_{k+1} = (k+1)(2(k+1) - 1) = 2k^2 +3k +1
    We also know that:
    \ S_{k+1} = S_k + u_{k+1}
    As
    u_{k+1} = 4(k+1) - 3 = 4k +1

 \Rightarrow S_{k+1} = 2k^2 - 1 + (4k+1) = 2k^2 + 3k +1
    Blah blah blah proved by induction etc.
    I can't do those for ****
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    (Original post by L'Evil Fish)
    I can't do those for ****
    Lol but it's so pretty
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    (Original post by joostan)
    Lol but it's so pretty
    If I did that, do you think I would have gotten marks anyway?
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    (Original post by L'Evil Fish)
    I can't do those for ****
    Isn't it just a simple process each time. Prove for base case, k, k+1.
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    (Original post by reubenkinara)
    Is it just a simple process each time. Prove for base case, k, k+1.
    Yeah, but it never works for me :9
 
 
 
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