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    Attachment 534779
    Attachment 534779534781

    how do you work out q 8 part iii) and also part iii) of the roots of polynomial question? (not the complex one)

    I don't fully understand the markscheme. So if anyone could show the root of polynomial part iii) step by step that would be wonderful

    For q 8...I got n/n-1 i believe ( i don't fully remember since I did it yesterday and i misplaced my working). But why is the answer 1- (n/n-1)?

    Again workibg out would be extremely helpful. Thank you again!!
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    Can't see what the pictures are, what year is it so o can look it up


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    I this off the January 2012 paper?

    The 1 comes from n/(n+1) tending to 1 as n tends to infinity. The -n/(n+1) comes from the fact that the sum to infinity is missing the first n terms. From this you get -n/(n+1) as this part isn't being cancelled out by an earlier term. Therefore you get 1- n/(n-1)
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    (Original post by drandy76)
    Can't see what the pictures are, what year is it so o can look it up


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    It's paper june 2012 (ocr non mei). I apologise for the inconvenience! Something must have went wrong while attatching on my phone...


    It is question 8 iii) and 6 part ii)
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    (Original post by jonnydowe)
    I this off the January 2012 paper?

    The 1 comes from n/(n+1) tending to 1 as n tends to infinity. The -n/(n+1) comes from the fact that the sum to infinity is missing the first n terms. From this you get -n/(n+1) as this part isn't being cancelled out by an earlier term. Therefore you get 1- n/(n-1)
    Oh I see thank you! It's the june 2012 paper ocr non mei. So when n tends to infinity the expression doesn't tend to 0 but 1? Just to confirm.
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    (Original post by JulieEdiz)
    Oh I see thank you! It's the june 2012 paper ocr non mei. So when n tends to infinity the expression doesn't tend to 0 but 1? Just to confirm.
    Yeah. Because at infinite the difference between n and n+1 is basically zero. Hence you get n/n which = 1
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    Should be noted that you can also use the equation from part i) to obtain 8/5 directly


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    (Original post by jonnydowe)
    Yeah. Because at infinite the difference between n and n+1 is basically zero. Hence you get n/n which = 1
    Sorry it is late but i just wanted to ask..

    sum of infinity from n+1= sum to infinity from 1 - sum from 1 to n...

    Therefore why isn't it 0- n/n+1...

    I understood what you said but it just doesn't make sense it i look at it from this point of view...
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    (Original post by JulieEdiz)
    Sorry it is late but i just wanted to ask..

    sum of infinity from n+1= sum to infinity from 1 - sum from 1 to n...

    Therefore why isn't it 0- n/n+1...

    I understood what you said but it just doesn't make sense it i look at it from this point of view...
    Sum to infinity from n+1 is all terms from n+1 to infinity, i.e. n+1, n+2, n+3

    Sum to infinity from 1 is 1, 2, 3, 4,5 etc including n+1 to infinity terms as they are inside of this region

    sum from 1 to n is all terms 1,2,3,4,5 to n-1, and n.

    so, from n+1 to infinity is sum from infinity from 1 minus the terms from 1 to n which you don't want, as you only want between n+1 and infinity
 
 
 
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