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    Why is:
    \displaystyle\sum_{r=1}^n 2 = 2n

    I understand this is equal to 2 + 2 + 2 + 2 + 2 ...... 2+2 but why is it 2n. Im doing method of differences. Cheers
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    Do you just want to understand it intuitively, or are you looking for more of a proof?
    If the latter, it's probably easiest to use induction.
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    when r=1 n=1 sum= 2
    When r=1 n=2 sum= 2+2 = 2(2)= 4
    When r=1 n=3 sum= 2+2+2= 3(2) = 6
    When r=1 n=n sum= n(2) = 2n
    Sorry that was the easiest way I could think to explain it?


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    (Original post by placenta medicae talpae)
    Do you just want to understand it intuitively, or are you looking for more of a proof?
    If the latter, it's probably easiest to use induction.
    I only wanted to understand it but, is the proof neccesary to learn?
    I'm self teaching so I havent covered proof by induction as of current.

    Also how does one read the notation out?
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    (Original post by UKBrah)
    Why is:
    \displaystyle\sum_{r=1}^n 2 = 2n

    I understand this is equal to 2 + 2 + 2 + 2 + 2 ...... 2+2 but why is it 2n. Im doing method of differences. Cheers
    It is 2n because there are n twos.

    The idea of proving this by induction is quite amusing.
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    Perhaps easier to understand: As multiplication is distributive in this case, you can take the 2 out of the sum giving

    \displaystyle\sum_{r=1}^n 2 = 2 \displaystyle\sum_{r=1}^n 1 = 2 \underbrace{(1+1+...+1)}_{n \text{ lots of 1}}=2n.

    I dunno if that's made it any clearer, but that's how I think of it!
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    (Original post by olipal)
    Perhaps easier to understand: As multiplication is distributive in this case, you can take the 2 out of the sum giving

    \displaystyle\sum_{r=1}^n 2 = 2 \displaystyle\sum_{r=1}^n 1 = 2 \underbrace{(1+1+...+1)}_{n \text{ lots of 1}}=2n.

    I dunno if that's made it any clearer, but that's how I think of it!
    Excellent! got it mate cheers.

    \displaystyle\sum_{r=1}^n k = kn
    Is that correct?
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    (Original post by UKBrah)
    Excellent! got it mate cheers.

    \displaystyle\sum_{r=1}^n k = kn
    Is that correct?
    Yep, provided it's not \displaystyle\sum_{k=1}^n k.
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    (Original post by olipal)
    Yep, provided it's not

    \displaystyle\sum_{k=1}^n k
    Sick, thanks alot mate. How do I read the sum notation out?
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    (Original post by olipal)
    Yep, provided it's not \displaystyle\sum_{k=1}^n k.
    Provided it is \displaystyle\sum_{k=1}^n k or it isnt?
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    (Original post by UKBrah)
    I only wanted to understand it but, is the proof neccesary to learn?
    I'm self teaching so I havent covered proof by induction as of current.

    Also how does one read the notation out?
    Oh right, cool.
    Are you teaching yourself AS level?
    If you're not doing further maths, you won't need to do much proving things.

    You can read the sum out as 'the sum from 1 to n of 2' and in other circumstances might need to add 'indexed by r' on the end.

    All you're doing is writing down a 2 for each of the numbers 1, 2, 3, ..., n, and adding them up.
    So the sum will be twice the number of 2's you write down, which is 2n.
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    (Original post by placenta medicae talpae)
    Oh right, cool.
    Are you teaching yourself AS level?
    If you're not doing further maths, you won't need to do much proving things.

    You can read the sum out as 'the sum from 1 to n of 2' and in other circumstances might need to add 'indexed by k' on the end.

    All you're doing is writing down a 2 for each of the numbers 1, 2, 3, ..., n, and adding them up.
    So the sum will be twice the number of 2's you write down, which is 2n.
    I'm self teaching AS further maths on MEI.
    Thanks
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    (Original post by UKBrah)
    I'm self teaching AS further maths on MEI.
    Thanks
    Oh right!
    For FP1 you will need to do induction, so once you've covered that, this question will probably come really naturally to you.
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    (Original post by placenta medicae talpae)
    Oh right!
    For FP1 you will need to do induction, so once you've covered that, this question will probably come really naturally to you.
    Oh right, so should I learn induction before method of differences?
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    \displaystyle\sum_{r=1}^n k = kn

    This is k+k+k...n times, so kn

    \displaystyle\sum_{k=1}^n k

    This is 1+2+3...+n which is n(n+1)/2

    Look at what notation is used below the summation point and it should be clear what the difference is between the two
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    Is the method of differences the thing where you consider the difference between successive terms in the sequence?

    Coz if so, I think it doesn't really add anything to the process or help you out.
    You already know that each term is 2 more than the last one because that's how it's constructed :hmmmm:
 
 
 
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