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Math Help - summation checkin

  1. #1
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    summation checkin

    If \sum^{n}_{r=1}U_r=3n^2+2n , find u .

    My working :

    From the result , i can see that it uses this formula

     \sum^{n}_{r=1}\frac{n(n+1)}{2} , so r exists.


    \frac{1}{2}xr^2+\frac{1}{2}xr-n=3n^2+2n

    By comparing , i got x=6 , and thus the answer is 6r-1 .

    I just want to know if there is a simpler approach to questions like this .
    Thanks .
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  2. #2
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    Hello, mathaddict!

    If \sum^{n}_{r=1}U_r\:=\:3n^2+2n , find U.

    Like you, I saw that the function is linear: . U_r \:=\:ar + b

    That is, we have: . \sum^n_{r=1}\,(ar + b) \;=\;3n^2 + 2n

    Then: . a\sum^n_{r=1}r + b\sum^n_{r=1}1 \;=\;3n^2+2n


    Knowing some summation formulas, we have: . a\cdot\frac{n(n+1)}{2} + b\cdot n \:=\:3n^2 + 2n

    . . which simplifies to: . \frac{a}{2}n^2 + \left(\frac{a}{2}+b\right)n \;=\;3n^2+2


    Equating coefficients, we have: . \begin{Bmatrix}\dfrac{a}{2} &=&3 \\ \\[-4mm] \dfrac{a}{2} + b &=& 2 \end{Bmatrix}

    . . Solve the system and get: . a = 6,\;b = -1


    Therefore: . U_r \:=\:6r-1

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  3. #3
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    THanks Soroban , your working is so much neater .
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  4. #4
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    Simpler :

    \sum^{n}_{r=1}U_r=3n^2+2n

    U_n = \sum^{n}_{r=1}U_r - \sum^{n-1}_{r=1}U_r

    U_n = 3n^2+2n-3(n-1)^2-2(n-1) = 6n-1
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