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    Question 3)a):
    http://www.toothill.notts.sch.uk/dat...W-QP-JUN05.PDF

    I have NO IDEA how to do it! I can do many sigma notation questions on other papers but this one, I don't know.

    Please help me! Thank you!!
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    (Original post by Konnichiwa)
    ...
    This may help:

    \displaystyle\sum_{i=1}^n [a(b+c)]=\sum_{i=1}^n[ab+ac]=\sum_{i=1}^n ab+\sum_{i=1}^n ac

    If you're still stuck, please clarify your problem.
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    I know that, but I am still stuck. I have done so many other similar questions, try it and you will know it's different from the others.

    If you can do it, please share with me how you did it.
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    (Original post by Konnichiwa)
    I know that, but I am still stuck. I have done so many other similar questions, try it and you will know it's different from the others.

    If you can do it, please share with me how you did it.
    It's no great difference to any other.

    Since you know the bits I mentioned, all you need to is subtract the RHS of one formula from the RHS of the other. Take out a factor of n(n+1)/12 and rearrange the rest.
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    Well it's difficult getting into the form they want, that's what it is different. I've tried doing that and factorising many times, yet no step further.

    It is harder than it seems.
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    (Original post by Konnichiwa)
    Well it's difficult getting into the form they want, that's what it is different. I've tried doing that and factorising many times, yet no step further.

    It is harder than it seems.
    It would have help if you'd said as such in your opening post.

    Taking out the factors I mentioned gives:

    \displaystyle\frac{n(n+1)}{12}[3n(n+1)-2(2n+1)]

    Now expand and then factorise what's in the square brackets.
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    (Original post by ghostwalker)
    It would have help if you'd said as such in your opening post.

    Taking out the factors I mentioned gives:

    \displaystyle\frac{n(n+1)}{12}[3n(n+1)-2(2n+1)]

    Now expand and then factorise what's in the square brackets.
    Thanks! I finally got it now, I can't believe I did not see that.
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    (Original post by Konnichiwa)
    Thanks! I finally got it now, I can't believe I did not see that.
    np.
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    hi I need help on this question please got exam on Friday Name:  paper question.png
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    (Original post by mathsRus)
    hi I need help on this question please got exam on Friday Name:  paper question.png
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    \displaystyle\sum_{r=1}^n (2r-1)^2=\sum_{r=1}^n (4r^2-4r+1)=4\sum_{r=1}^n r^2- 4\sum_{r=1}^n r+ \sum_{r=1}^n 1

    Does that help?

    If you are really stuck, I suggest you go learn the chapter again.
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    O I am sorry I did not specify. I needed help on part b of the question where I did use the answer from previous part but I don't know what put in place of (n)

    Thanks for your reply.
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    (Original post by mathsRus)
    O I am sorry I did not specify. I needed help on part b of the question where I did use the answer from previous part but I don't know what put in place of (n)

    Thanks for your reply.
    The smallest odd number after 100 is 101, so what value of n gives 101?

    Similarly for largest odd number before 200

    You now have the limits on your summation.
 
 
 
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