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Math Help - limit series question..

  1. #1
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    limit series question..

    <br />
\lim _{x->\infty}\frac{1^2+3^2+..+(2n-1)^2}{2^2+4^2+..+(2n)^2}<br />

    i wanted to make a total sum of each series but there is no constant gap
    between two members 1 4 9 ..
    i cant make an expression of a sum
    for the numerator and denominator

    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    <br />
\lim _{x->\infty}\frac{1^2+3^2+..+(2n-1)^2}{2^2+4^2+..+(2n)^2}<br />

    i wanted to make a total sum of each series but there is no constant gap
    between two members 1 4 9 ..
    i cant make an expression of a sum
    for the numerator and denominator

    ??
    Using standard formulae and a little bit of basic algebra:

    \sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}.


    \sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = \frac{2n(n+1)(2n+1)}{3}.
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  3. #3
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    lt (n tending to infinity)[{1^2+2^2+3^2+........+(2n)^2}/{2^2+4^2+......+(2n)^2}-1]

    now,1^2+2^2+3^2+......+(2n)^2=2n*(2n+1)*(4n+1)/6.

    &,2^2+4^2+......+(2n)^2=2^2{1^2+2^2+3^2+.....+n ^2}
    =2^2*n*(n+1)*(2n+1)/6

    hence,lt(n--->infinity)[(2*2*4)/(4*1*2)-1]

    =1.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Using standard formulae and a little bit of basic algebra:

    \sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}.


    \sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = \frac{2n(n+1)(2n+1)}{3}.
    i didnt study this form can you explain this in simpler terms??

    without the \Sigma
    i understand this but i need to solve such thing in a test
    and will not be accepted
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  5. #5
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    Quote Originally Posted by transgalactic View Post

    i understand this but i need to solve such thing in a test
    and will not be accepted
    I'm sorry but this is false.

    Notation \sum is very useful and I think you should take advantage of.

    Having \sum_{k=1}^3k=1+2+3, and that's all. - Basic formulae could be found in Google.
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    i didnt study this form can you explain this in simpler terms??

    without the \Sigma
    i understand this but i need to solve such thing in a test
    and will not be accepted
    Quote Originally Posted by transgalactic View Post
    <br />
\lim _{x->\infty}\frac{1^2+3^2+..+(2n-1)^2}{2^2+4^2+..+(2n)^2}<br />

    i wanted to make a total sum of each series but there is no constant gap
    between two members 1 4 9 ..
    i cant make an expression of a sum
    for the numerator and denominator

    ??
    I, clearly wrongly, assumed from this that you were familiar with the results for \sum_{k=1}^n k and \sum_{k=1}^n k^2 and it was simply the 'constant gap' that was causing the problem.

    If that's not the case, then I don't see how you could do the question, in which case I don't see why you would be given it in the first place.
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  7. #7
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    i can understand how you get this transition

    <br />
 4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}<br />
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    i can understand how you get this transition

    <br />
4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}<br />
    So what's your question?
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  9. #9
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    there are sums of n members +n members
    how you get 4 members on the right side.
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    there are sums of n members +n members
    how you get 4 members on the right side.
    Standard results readily found using Google:

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

    \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

    It is basic algebra (or should be) to substitute these expressions into 4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n and get \frac{n(2n-1)(2n+1)}{3}.
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  11. #11
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    i cant see a pattern in your formulas which were found in google
    ??

    i know that the sum of number
    from 1 to n

    is (n+1)*n/2 so the sum of k is correct

    but the k^2 case doesnt hold for it
    ??
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  12. #12
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    Quote Originally Posted by transgalactic View Post
    i cant see a pattern in your formulas which were found in google
    ??

    i know that the sum of number
    from 1 to n

    is (n+1)*n/2 so the sum of k is correct

    but the k^2 case doesnt hold for it
    ??
    The formula I have given for the sum of the square of the first n integers is correct. Are you now trying to derive that formula? I really don't understand what your problem is now .... As far as I can see the question you originally posted has been answered ....
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