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    Merry Christmas everyone,

    I'm just a little confused on some questions that frequently pop up in the papers.

    Thay say something like,

    Prove that the sum of the first n positive integers is given by
    1/2 n(n + 1)

    And I know that you show something like

    Sn= 1+2+3+...+n
    Sn= n+(n-1)+(n-2)+...+1
    2Sn= n(n+1)
    Sn= 1/2 n(n+1)

    But it makes no sense at all :s

    Why does this happen and why does it become to that formula?
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    Sn = n + (n-1) + (n-2) + ... + 1
    Rearrange the terms into pairs that each total n+1:
    Sn = [n + 1] + [(n-1) + 2] + [(n-2) + 3] + ...
    There will always be n/2 such pairs in the sequence, so:
    Sn = n/2(n+1)
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    S_n = 1 + 2 + ... + n should be obvious

    S_n = n + (n-1) + (n-2) + ... + 1 is S_n written backwards (e.g. 1 + 2 + 3 + 4 + 5 = 5 + 4 + 3 + 2 + 1, they're the same thing)

    Now if you add the first line and the second line, on the left hand side you get S_n + S_n = 2S_n.

    On the right hand side, add the terms like they're in a column.

    So first add the 1 and the n, to get 1 + n
    Then add the 2 + (n-1), to get 1 + n
    Then the 3 + (n-2), to get 1 + n
    ...
    Then the n + 1, to get 1 + n

    You get 1 + n every time and there are n terms, so all together you have n(1+ n) on the right hand side.

    Left hand side = 2S_n
    Right hand side = n(n+1)

    So 2S_n = n(n+1)
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      (Original post by Swayum)
      S_n = 1 + 2 + ... + n should be obvious

      S_n = n + (n-1) + (n-2) + ... + 1 is S_n written backwards (e.g. 1 + 2 + 3 + 4 + 5 = 5 + 4 + 3 + 2 + 1, they're the same thing)

      Now if you add the first line and the second line, on the left hand side you get S_n + S_n = 2S_n.

      On the right hand side, add the terms like they're in a column.

      So first add the 1 and the n, to get 1 + n
      Then add the 2 + (n-1), to get 1 + n
      Then the 3 + (n-2), to get 1 + n
      ...
      Then the n + 1, to get 1 + n

      You get 1 + n every time and there are n terms, so all together you have n(1+ n) on the right hand side.

      Left hand side = 2S_n
      Right hand side = n(n+1)

      So 2S_n = n(n+1)
      Wait, if n = 1, why does n - 1 = 2?

      Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

      :confused:
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      (Original post by im so academic)
      Wait, if n = 1, why does n - 1 = 2?

      Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

      :confused:
      Given a=1 in this case, and d=1 we can simplify your formula to the above.
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        (Original post by Circadian_Rhythm)
        Given a=1 in this case, and d=1 we can simplify your formula to the above.
        Oh, thank you very much!

        But is it possible to do it the "[2a + (n-1)d]" way and then simplify it, right?
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        (Original post by im so academic)
        Wait, if n = 1, why does n - 1 = 2?

        Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

        :confused:
        If n = 1, you're asking to sum the series with just one term: 1. So there doesn't exist an (n-1)th term in this context. The question is to sum

        1 + 2 + 3 + ... + n

        You can let n = 3, in which case you're summing 1 + 2 + 3 = 6. The way our method works is that

        S_3 = 1 + 2 + 3 (forwards)
        S_3 = 3 + 2 + 1 (backwards)

        Add those lines to get 2S_3 = (1+3) + (2+2) + (3+1) = 4 + 4 + 4 = 3*4 = 12
        So S_3 = 6

        You can let n = 1, in which case you're summing 1 = 1. The way our methods works is that

        S_1 = 1 (forwards)
        S_1 = 1 (backwards)

        Add and you get 2S_1 = 1 + 1 = 2

        So S_1 = 1

        S_n = 0.5n(2a + d(n-1)) is the sum starting from the number a and with any difference between two terms. The person above asked for the formula starting from the number 1 and with a difference of 1 between any two numbers.

        Clearly, starting at a with a difference of d is a more general way of doing it, and that formula is derived in exactly the same way as writing the series backwards and adding it.
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        (Original post by im so academic)
        Oh, thank you very much!

        But is it possible to do it the "[2a + (n-1)d]" way and then simplify it, right?
        Yes, just scribbled it down.

        Sn=n/2[2a+d(n-1)]

        We know a=1 and d=1

        Sn=n/2[(2x1)+1(n-1)]

        Sn=n/2[(2+n-1)]

        Sn=n/2(n+1)

        You could then multiply both sides by two and then divide again, but its trivial. And that's the 'proof'. Enjoy
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        (Original post by im so academic)
        Wait, if n = 1, why does n - 1 = 2?

        Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

        :confused:
        Oh I see what you've misunderstood. I said that

        "S_n = 1 + 2 + ... + n should be obvious

        S_n = n + (n-1) + (n-2) + ... + 1 is S_n written backwards"

        We're not saying that the terms that are above and below each other are EQUAL. No no no. We're saying, let's add them TOGETHER. It's like

        1 + 2 + 3 + 4 is still the same as 4 + 3 + 2 + 1 but written backwards, isn't it? The quantity doesn't change.

        But the point is to now add the first terms together: 1 and 4

        Then add second terms together: 2 and 3

        Then third terms: 3 and 2

        Then fourth terms: 4 and 1

        Notice all of these are 5! 1 + 4 = 5, 2 + 3 = 5, 3 + 2 = 5, 4 + 1 = 5. That's the point - they all equal the same thing, and there are n of them, so we can add them together easily.
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        (Original post by Circadian_Rhythm)
        Yes, just scribbled it down.

        Sn=n/2[2a+d(n-1)]

        We know a=1 and d=1

        Sn=n/2[(2x1)+1(n-1)]

        Sn=n/2[(2+n-1)]

        Sn=n/2(n+1)

        You could then multiply both sides by two and then divide again, but its trivial. And that's the 'proof'. Enjoy
        In this question, they wouldn't accept that as a proof because you haven't explained why S_n = 0.5n(2a + d(n-1)). Usually, you can just quote that formula in C1 questions, but you wouldn't be allowed to/it wouldn't be acceptable in this question.
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        (Original post by Swayum)
        In this question, they wouldn't accept that as a proof because you haven't explained why S_n = 0.5n(2a + d(n-1)). Usually, you can just quote that formula in C1 questions, but you wouldn't be allowed to/it wouldn't be acceptable in this question.
        Yes, fair enough, I see that, but academic was asking why the other formula couldn't be used I think. Hopefully I don't get this question in January, hopefully something more exciting...
       
       
       
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