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Math Help - pre calculus involving limits only

  1. #1
    Member grgrsanjay's Avatar
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    pre calculus involving limits only

    let \alpha = \displaystyle\lim_{n\to\infty} \left(\frac{1^2 + 2^2 + ..... +n^2}{n^3}\right) and
    \beta = \displaystyle\lim_{n\to\infty} \left(\frac{(1^3 - 1^2 )+(2^3 -2^2 )+ ..... +(n^3 - n^2 )}{n^4}\right) ,then which of the following is correct
    (A) \alpha = \beta
    (B) \alpha < \beta
    (C) 4\alpha - 3\beta =0
    (D) 3\alpha - 4\beta =0
    Last edited by grgrsanjay; August 31st 2010 at 07:30 AM.
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  2. #2
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    Quote Originally Posted by grgrsanjay View Post
    let \alpha = \displaystyle\lim_{n\to\infty} \left(\frac{1^2 + 2^2 + ..... +n^2}{n^3}\right) and
    \beta = \displaystyle\lim_{n\to\infty} \left(\frac{(1^3 - 1^2 )+(2^3 -2^2 )+ ..... +(n^3 - n^2 )}{n^4}\right) ,then which of the following is correct
    (A) \alpha = \beta
    (B) \alpha < \beta
    (C) 4\alpha - 3\beta =0
    (D) 3\alpha - 4\beta =0
    \alpha contains a "sum of squares" in the numerator.

    \beta contains the difference between a "sum of cubes" and a "sum of squares" in the numerator.

    When you express those sums in "closed form", you can evaluate the limits.

    You will find the correct answer is one of the 4 listed.
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  3. #3
    Member grgrsanjay's Avatar
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    Quote Originally Posted by Soroban
    Hello, grgrsanjay!

    Archie Meade gave you an excellent game plan . . . Could you follow it?


    We are expected to know these summation formulas:

    . . 1^2 + 2^2 + 3^2 + \hdots + n^2 \:=\:\dfrac{n(n+1)(2n+1)}{6}

    . . 1^3 + 2^3 + 3^3 + \hdots + n^3 \:=\:\dfrac{n^2(n+1)^2}{2}


    \displaystyle \text{Let }\:\alpha \:=\:\lim_{n\to\infty} \left(\frac{1^2 + 2^2 + ..... +n^2}{n^3}\right) \;\;[1]

    \displaystyle \text{Let }\:\beta \:=\: \lim_{n\to\infty} \left(\frac{(1^3 - 1^2 )+(2^3 -2^2 )+ \hdots +(n^3 - n^2 )}{n^4}\right) \;\;[2]


    \text{Then which of the following is correct?}

    . . (A)\; \alpha \:=\: \beta\qquad (B)\;\alpha \:<\: \beta \qquad (C)\; 4\alpha - 3\beta \:=\:0 \qquad (D)\;3\alpha - 4\beta =0


    In [1], we have: . \dfrac{1^2+2^2+3^2+\hdots+n^2}{n^3} \;=\;\dfrac{\frac{n(n+1)(2n+1)}{6}}{n^3} \;=\;\dfrac{2n^3+3n^2+n^2}{6n^3}\;=\;\dfrac{1}{3} + \dfrac{1}{2n} + \dfrac{1}{6n^2}

    . . Hence: . \displaystyle \alpha \;=\;\lim_{n\to\infty}\left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right) \;=\;\frac{1}{3}



    In [2], we have: . \dfrac{\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}}{n^4} \;=\;\dfrac{3n^4 + 2n^3 - 3n^2 - 2n}{12n^4} \;=\;\dfrac{1}{4} + \dfrac{1}{6n} - \dfrac{1}{4n^2} - \dfrac{1}{6n^3}


    Hence: . \displaystyle \beta \;=\;\lim_{n\to\infty}\left(\frac{1}{4} + \frac{1}{6n} - \frac{1}{4n^2} - \frac{1}{6n^3}\right) \;=\;\frac{1}{4}


    The correct answer is (D).


    yea i could follow it
    it was very easy but i dint use the logic properly
    thank you sorobon
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