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    Use formulae for \displaystyle\sum_{i=1}^n i^2

    \displaystyle\sum_{i=1}^n i and \displaystyle\sum_{i=1}^n 1

    To find the following: \displaystyle\sum_{i=1}^n (2r+1)

    I'm not really too sure how to go about it.
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    Well, you need to use standard rules to find it, can't you just deduce them?

    For example, what is \sum_{i=1}^{n} 1

    It will be more rewarding if you try to find them yourself, rather than me to give you a list of manipulations and standard formulae.
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    (Original post by gff)
    Well, you need to use standard rules to find it, can't you just deduce them?

    For example, what is \sum_{i=1}^{n} 1

    It will be more rewarding if you try to find them yourself, rather than me to give you a list of manipulations and standard formulae.
    That would just be n? Ohhhhhh i think i get it now!

    So for \displaystyle\sum_{i=1}^n (2r+1) It would be

    2(\frac{1}{2}n(n+1))+1=n(n+2)?
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    Wait, don't hurry that much.

    First, is that r, or is this meant to be i?

    Look at the question, what do you think you can break this sum into? (Perhaps two sums? What are they?)

    Edit:

    I didn't read the answer before posting. You got it right, but did you use what was required? Can't see from here.
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    (Original post by gff)
    Wait, don't hurry that much.

    First, is that r, or is this meant to be i?

    Look at the question, what do you think you can break this sum into? (Perhaps two sums? What are they?)

    Edit:

    I didn't read the answer before posting. You got it right, but did you use what was required? Can't see from here.
    ye that r was meant to be an i sorry.

    I think so I did this: \displaystyle\sum_{i=1}^n (2i+1)= \displaystyle\sum_{i=1}^n 2i + \displaystyle\sum_{i=1}^n 1

    =2(\frac{1}{2}n(n+1))+n=n(n+2) Although looking at that i don't get how the last step would work
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    What is the last step?

    \sum_{i=1}^{n} (2i + 1) = \sum_{i=1}^{n} 2i + \sum_{i=1}^{n} 1 = 2 \sum_{i=1}^{n} i + n

     = 2(\frac{1}{2} n(n + 1)) + n = n(n + 1) + n = n^2 + n + n = n^2 + 2n = n(n + 2)

    Also here are some identities and manipulations.

    Bla, bla.. why did I write that much? (rofl)
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    (Original post by gff)
    What is the last step?

    \sum_{i=1}^{n} (2i + 1) = \sum_{i=1}^{n} 2i + \sum_{i=1}^{n} 1 = 2 \sum_{i=1}^{n} i + n

     = 2(\frac{1}{2} n(n + 1)) + n = n(n + 1) + n = n^2 + n + n = n^2 + 2n = n(n + 2)

    Also here are some identities and manipulations.

    Bla, bla.. why did I write that much? (rofl)
    Ah of course , feel silly didnt spot that haha. Cheers for the help.
 
 
 

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