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Proof for nth term formula watch

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    Im really desperate
    does anyone know the proof for the following nth term formula when the difference is changing.

    a(n-1)d+1/2(n-1)(n-2)c

    This is a formula from the cgp higher maths GCSE book-but i need to find a proof to show why it works. If anyone knows a proof or can work one out could they plz plz post it clearly in stages.

    Thanks
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    (Original post by aacharyauk)
    Im really desperate
    does anyone know the proof for the following nth term formula when the difference is changing.

    a(n-1)d+1/2(n-1)(n-2)c

    This is a formula from the cgp higher maths GCSE book-but i need to find a proof to show why it works. If anyone knows a proof or can work one out could they plz plz post it clearly in stages.

    Thanks
    aaaarghhhhhh maths!!
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    (Original post by rednirt)
    aaaarghhhhhh maths!!
    4th year Cambridge Maths Degree Level!!
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    (Original post by bono)
    4th year Cambridge Maths Degree Level!!
    bono-plz plz what do u mean-plz reply back quickly. Can u give me a proof. What do u mean by cambridge-i need is desperately
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    What does this formula represent?
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    (Original post by theone)
    What does this formula represent?
    a is the first term
    d is the first difference in the sequence
    c is the changing difference from one difference to the next
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    look if anyone is out there and knows could u reply in the next few mins cos i am seriously desperate for a proof of this formula. It will be the end of two weeks worth of struggle.
    PLZ PLZ PLZ PLZ
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    (Original post by aacharyauk)
    a is the first term
    d is the first difference in the sequence
    c is the changing difference from one difference to the next

    is this thingy for A level or degree maths ???
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    (Original post by aacharyauk)
    Im really desperate
    does anyone know the proof for the following nth term formula when the difference is changing.

    a(n-1)d+1/2(n-1)(n-2)c

    This is a formula from the cgp higher maths GCSE book-but i need to find a proof to show why it works. If anyone knows a proof or can work one out could they plz plz post it clearly in stages.

    Thanks
    This is called a geometric sequence:

    Let S, be the sum of a sequence...

    (S = 1 + 2 + 4 + 8 + 16.)

    Times this by 2:

    (2S = 2 + 4 + 8 + 16 + 32)

    Notice that both have 2, 4, 8, 16 in common, and the sum of this is S-1 rom teh first equation, and 2S-32 from the second. So:

    S - 1 = 2S - 32, giving S=31

    This can be used to find the sum of any geomeric sequence. Let S be the sum of the first n term of the sequence. Then:

    S = a + ar + ar^2 + ... + ar^n-2 + ar^n-1.

    If you multiply is by r, you get:

    Sr = ar + ar^2 + + ar^3 + ... ar^n-1 + ar^n

    The right sides in these 2 equations have the terms ar + ar^2 + ... ar^n-2 + ar^n-1 in common, so:

    S-a = ar + ar^2 + ... ar^n-2 + ar^n-1 = Sr - ar^n

    Which gives:

    S(1-r) = a(1-r^n)

    And so S = [a(1-r^n)]/[1-r]

    Notice that r can both be positive and negative for this to work.
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    (Original post by 2776 2)
    This is called a geometric sequence:
    Actually this is not a geometric series since the difference between terms does not increase exponentially.

    For instance the series 1,2,4,7,11,16 has a=1, b=1 and c=1 but is not geometric.
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    (Original post by theone)
    Actually this is not a geometric series since the difference between terms does not increase exponentially.

    For instance the series 1,2,4,7,11,16 has a=1, b=1 and c=1 but is not geometric.
    I just realised that.
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    (Original post by 2776 2)
    This is called a geometric sequence:

    Let S, be the sum of a sequence...

    (S = 1 + 2 + 4 + 8 + 16.)

    Times this by 2:

    (2S = 2 + 4 + 8 + 16 + 32)

    Notice that both have 2, 4, 8, 16 in common, and the sum of this is S-1 rom teh first equation, and 2S-32 from the second. So:

    S - 1 = 2S - 32, giving S=31

    This can be used to find the sum of any geomeric sequence. Let S be the sum of the first n term of the sequence. Then:

    S = a + ar + ar^2 + ... + ar^n-2 + ar^n-1.

    If you multiply is by r, you get:

    Sr = ar + ar^2 + + ar^3 + ... ar^n-1 + ar^n

    The right sides in these 2 equations have the terms ar + ar^2 + ... ar^n-2 + ar^n-1 in common, so:

    S-a = ar + ar^2 + ... ar^n-2 + ar^n-1 = Sr - ar^n

    Which gives:

    S(1-r) = a(1-r^n)

    And so S = [a(1-r^n)]/[1-r]

    Notice that r can both be positive and negative for this to work.
    Is this proof for the formula a(n-1)d+1/2(n-1)(n-2)c- cos i am doing GCSE maths coursework for squares in rectangles and the thing is that i have found an algebraic formula but i have used this formula to get it. Now this formula is from a textbook and is not my work. My teacher said i have to prove this. So if anyone knows a proof for this formula-ie in words or algebra(preferably) of why it gives you the nth term then i can include this in my coursework to get higher marks. I need it quickly cos i gotta hand it in tommorrow. THANKS u guys for trying to help. I appreciate it.
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    First let us consider the simpler series:

    a, a+b, a+2b... a+(n-1)b. Let the sum of this be S. Now writing this backwards, S = a+(n-1)b + ... a+b + a. Now adding corresponding terms. 2S = n(2a + (n-1)b) so S = n/2(2a+(n-1)b).

    Now let us consider your series, defined as:

    a, a+d, a+2d+c, a+3d+2c ... a+(n-1)d+(n-2)c.

    Now this is equal to:

    a + a+d + ... a+(n-1)d = n/2(2a+(n-1)d). Also c + 2c + ... (n-2)c = (n-2)(n-1)c/2.

    Adding gives an + n(n-1)d/2 + (n-2)(n-1)c/2.

    This isn't the same as what you initally posted, but I think this is right. Hope this helps.

    Is the series a, a+d, a+2d+c, a+3d+2c ... a+(n-1)d+(n-2)c. what you actually mean?
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    (Original post by theone)
    First let us consider the simpler series:

    a, a+b, a+2b... a+(n-1)b. Let the sum of this be S. Now writing this backwards, S = a+(n-1)b + ... a+b + a. Now adding corresponding terms. 2S = n(2a + (n-1)b) so S = n/2(2a+(n-1)b).

    <editing>
    what do u mean-i am really thick at maths. Sorry but seriously all i am going to do is copy your proof. Ill try and understand it tho.
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    (Original post by aacharyauk)
    what do u mean-i am really thick at maths. Sorry but seriously all i am going to do is copy your proof. Ill try and understand it tho.
    You are supposed to work this out yourself. Not just copy others work.
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    (Original post by aacharyauk)
    what do u mean-i am really thick at maths. Sorry but seriously all i am going to do is copy your proof. Ill try and understand it tho.
    U see the formula i posted is the one for working out the nth term. Now i need to put this formula in my work but cos its from a book i cant just put it in. I need to prove it and say why it works. Algebraically
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    (Original post by aacharyauk)
    U see the formula i posted is the one for working out the nth term. Now i need to put this formula in my work but cos its from a book i cant just put it in. I need to prove it and say why it works. Algebraically
    I know- i never copy anyones work but im in a real panic that why! Ill try and understand it tho.
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    Can you explain what kind of series this is? And i mean explain it properly.
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    (Original post by prince_capri)
    is this thingy for A level or degree maths ???
    this is a GCSE coursework
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    (Original post by theone)
    Can you explain what kind of series this is? And i mean explain it properly.
    OK -im really sorry. Right ummmmm
    I used this formula to get the nth term of this series

    1,3,6,10,15

    here i had to look at the third difference to get a constant and hence the formula became
    a + (n-1)d + (1/2)(n-1)(n-2)c1 + 1/6 (n-1)(n-2)(n-3)

    and for this series

    1, 5, 14, 30, 55

    the formula remained as it was as at the second difference i had a constant.

    Now i had to prove why this formula works for working out the nth term.
 
 
 
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