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    I understand how to expand r squared and 2r.

    But I'm stuck on 2n + 1.

    In the mark scheme they have done this: 2n(n+1) and (n+1) so it looks like 2n and 1 have been multiplied by (n+1)...could someone explain the understanding behind this please?
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    (Original post by Ravster)
    In the mark scheme they have done this: 2n(n+1) and (n+1) so it looks like 2n and 1 have been multiplied by (n+1)...could someone explain the understanding behind this please?
    This is the 6th time I'm answering this question on TSR, seems like a good question! :lol:

    I'll give you two ways to look at it:

    1. \displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)

    2. \displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}.

    Since in this case you have \sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

    Then you have f(r) = 1 being added together n+1 times to get a total of (2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1).
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    What paper year is this? I've never seen anything like this before, looks hard!
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    (Original post by 16characterlimit)
    What paper year is this? I've never seen anything like this before, looks hard!
    It's not any different to a normal series question.
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    (Original post by 16characterlimit)
    What paper year is this? I've never seen anything like this before, looks hard!
    June 2013 (R) and you're actually quite right. This was the only question of its kind in all the past papers, in the sense that it started from r=0, not hard, just different to the normal questions
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    (Original post by Zacken)
    It's not any different to a normal series question.
    But you are summing r normally, this one you have to sum n.

    edit: NEVER MIND, I haven't done FP1 for a while.

    edit2: no I am confused now, the standard sums only apply for the variable below the sum, r in this case. Help?
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    (Original post by 16characterlimit)
    But you are summing r normally, this one you have to sum n.

    edit: NEVER MIND, I haven't done FP1 for a while.

    edit2: no I am confused now, the standard sums only apply for the variable below the sum, r in this case. Help?
    Look at my post, n is just some random constant.

    So \sum_{r=1}^n (2n+1) = (2n+1)\sum_{r=1}^n (1) and you can't really get a simpler sum than just summing 1's up...
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    (Original post by Zacken)
    Look at my post, n is just some random constant.

    So \sum_{r=1}^n (2n+1) = (2n+1)\sum_{r=1}^n (1) and you can't really get a simpler sum than just summing 1's up...
    Oh right, thanks crises averted.

    Isn't a bit strange using the same letter for the constant as in the sum, like saying differentiate x + x, where the second x is a constant? Or am I missing something important.
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    (Original post by 16characterlimit)
    Oh right, thanks crises averted.

    Isn't a bit strange using the same letter for the constant as in the sum, like saying differentiate x + x, where the second x is a constant? Or am I missing something important.
    No, it's not. It's like you saying, "heeey! You're not allowed to do \sum_{r=1}^5 (5) because the number at the top and in the sum are the same wtf?!??!?!?!?!/!?!?". n is a fixed unchanging constant. r is the only thing that's changing.
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    (Original post by Zacken)
    No, it's not. It's like you saying, "heeey! You're not allowed to do \sum_{r=1}^5 (5) because the number at the top and in the sum are the same wtf?!??!?!?!?!/!?!?". n is a fixed unchanging constant. r is the only thing that's changing.
    thx +rep
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    (Original post by 16characterlimit)
    thx +rep
    No prob.
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    What exam board and year paper is this question from?
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    (Original post by Bobunitrd)
    What exam board and year paper is this question from?
    Edexcel June 2013 (R), as mentioned above in the thread.
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    (Original post by Zacken)
    Edexcel June 2013 (R), as mentioned above in the thread.
    Oh sorry didn't see that, thanks anyway
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    (Original post by Bobunitrd)
    Oh sorry didn't see that, thanks anyway
    No problem.
 
 
 
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