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Thread: Sequences

  1. #1
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    Sequences

    Evaluate 1 x 2^2 + 3 x 4^2 + 5 x 6^2 +... to n terms
    I have been using (2n-1)(2n^2) as my nth term. After finding my sum to n terms, i have been substituting a value in, continuously it is not coming out as intended. Possibly my nth term is incorrect. I have been using the rth term expressed as a polynomial in r method.
    Thanks.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Webby View Post
    Evaluate 1 x 2^2 + 3 x 4^2 + 5 x 6^2 +... to n terms
    I have been using (2n-1)(2n^2) as my nth term. After finding my sum to n terms, i have been substituting a value in, continuously it is not coming out as intended. Possibly my nth term is incorrect. I have been using the rth term expressed as a polynomial in r method.
    Thanks.
    Your T_n will be = (2n-1)(4n^2)

    $\displaystyle T_n = (2n-1)(2n)^2 $

    $\displaystyle S_n = \sum_{r=1}^x{(8n^3-4n^2)}$

    I don't think you need help from here, ask it if you feel so

    ------------------------------------
    EDIT:

    I will go ahead now

    $\displaystyle S_n = \sum_{r=1}^x{(8n^3-4n^2)}$

    $\displaystyle = \frac{8(x(x+1))^2}{4} - 4 \frac{(x(x+1)(2x+1))}{6} $

    $\displaystyle = 2(x(x+1))^2 - 2 \frac{(x(x+1)(2x+1))}{3} $

    Thus only simplification is left
    Last edited by ADARSH; Mar 18th 2009 at 03:12 AM. Reason: typo
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  3. #3
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    The sum of 4n^2, would be 4n(n+1)(2n+1)/6 if i am not mistaken.
    However i am coming up with a sum to n terms of 2n(n+1)(3n^2+n-1)/3. When substituting values n>1. I am coming out 4 ahead of what i should be. Not sure where i am going wrong.
    Last edited by Webby; Mar 18th 2009 at 03:18 AM.
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Webby View Post
    The sum of 4n^2, would be 4n(n+1)(2n+1)/6 if i am not mistaken.
    However i am coming up with a sum to n terms of 2n(n+1)(3n^2+n-1)/3. When substituting values n>1. I am coming out 4 ahead of what i should be.

    I had an idiotic typo , I have corrected it now



    As far as answer is concerned

    Sum of x terms $\displaystyle = 2(x(x+1))^2 - 2 \frac{(x(x+1)(2x+1))}{3}$

    Put 2 , sum should come out to be = 4+48 =52

    So lets put x= 2 in the answer

    $\displaystyle 2(2(3))^2 - 2 \frac{(2(3)(5))}{3} $

    $\displaystyle = 2(4*9) - 2 \frac{30}{3} $

    $\displaystyle = 72 - 20 $

    $\displaystyle = 52 $
    --------------------------------------
    Your mistake was that you considered the nth term

    = (2n-1)(2n^2)

    However it should be = (2n-1)(2n)^2
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  5. #5
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    Hello, Webby!

    Evaluate: .$\displaystyle 1\cdot2^2 + 3\cdot4^2 + 5\cdot6^2 + \hdots$ .to $\displaystyle n$ terms.

    $\displaystyle \text{I have been using }(2n-1)\underbrace{(2n^2)}_{{\color{red}\text{no}}}\tex t{ as my }n^{th}\text{ term.}$

    As you suspected, your $\displaystyle n^{th}$ term is wrong.
    We want the square of $\displaystyle {\color{red}2}n\!:\;\;(2n)^2 = 4n^2$


    ARDASH's work is excellent and correct!

    To simplify that formula: .$\displaystyle S_n\;=\;2n^2(n+1)^2 - \tfrac{2}{3}n(n+1)(2n+1)$

    . . Factor! . $\displaystyle \frac{2}{3}n(n+1)\bigg[3n(n+1) - (2n+1)\bigg]$

    Therefore: .$\displaystyle S_n \;=\;\frac{2}{3}n(n+1)(3n^2+n-1) $

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