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Math Help - Odd Sequence

  1. #1
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    Odd Sequence

    A sequence of odd numbers is grouped the following way;

    \{1\},\:\:\{3,5,7\},\:\:\{9,11,13,15,17\},\:\:\{19  ,21,23,25,27,29,31\},\:\:...

    How many numbers are there in the nth group?

    (this seemed straightforward enough. By observation; 2n-1)

    How many numbers are there from the 1st group to the (n-1)th group?

    (this is where everything goes downhill for me, would be great to get some help...)

    What is the first number in the nth group?

    What is the sum of all the numbers in the nth group?

    What is the sum of all the numbers in the 10th group?

    Once again, any help appreciated.
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  2. #2
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    Hello,

    You can have a look here : http://www.mathhelpforum.com/math-he...h-bracket.html
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  3. #3
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    Dear davidman,

    Number of elements in the n^{th} set= 2n-1

    Number of elements from the first set to the (n-1)^{th} set= \sum_{r=1}^{n-1}(2r-1)=n^2-2n+1

    First element in the n^{th} set= (n^2-2n+2)^{th} element=1+2(n^2-2n+2-1)=2n^{2}-4n+3

    Sum of all the elements from the first set to the n^{th} set= \frac{n^2}{2}[2+2(n^2-1)]=n^4

    Sum of all the elements from the first set to the (n-1)^{th} set= \frac{(n^2-2n+1)}{2}[2+2(n^2-2n+1-1)]=(n^2-2n+1)^2

    Sum of all the elements in the n^{th} set= n^{4}-(n^2-2n+1)^2

    Sum of all the elements in the 10^{th} set= 10^4-(10^2-20+1)^2= 3439

    Hope this helps.
    Last edited by Sudharaka; January 27th 2010 at 05:18 AM.
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  4. #4
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    I translated this problem from my book myself, so I might not be completely accurate with what the original problem is. But here's the answers as given in the back of my book.

    2n-1

    (n-1)^2

    2n^2-4n+3

    (2n-1)(2n^2-2n+1)

    3439
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  5. #5
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    Dear davidman,

    Of course I had miscalculated the last part which I corrected now.
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