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Math Help - Progression

  1. #1
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    Progression

    Need help with number (9).. Progression-dsc00560.jpg Thanks...
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  2. #2
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    If S_n=3n^2+4n, then V_n=S_n-S_{n-1}.
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  3. #3
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    Hello, MichaelLight!

    \displaystyle \text{9. If }\sum^n_{r=1}V_r \,=\,3n^2 + 4n,\,\text{find the }n^{th}\text{ term of the series.}

    \text{Hence, show that the series is an arithmetic series.}

    Since the sum of the terms is a quadratic, the sequence must be linear.

    The general term would be: / V_r \:=\:ar + b

    Then: . \displaystyle \sum^n_{r=1}V_r \;=\;\sum^n_{r=1}(ar + b) \;=\;a\sum^n_{r=1}r + b\sum^n_{r=1}1

    . . . =\;a\frac{n(n+1)}{2} + bn \;=\;\frac{a}{2}n^2 + \frac{a+2b}{2}n \;\;\text{ which equals }3n^2 + 4n


    So we have: . \begin{Bmatrix} \frac{a}{2} \;=\;3 \\ \\[-3mm]  \frac{a+2b}{2} \;=\;4\end{Bmatrix} \quad\Rightarrow\quad a =6,\;b=1

    . . Therefore: . V_r \:=\:6r + 1



    The sequence is: . 7, 13, 19, 25, 31\;\hdots . an arithmetic series

    . . with first term a = 7 and common difference d = 6.

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