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Math Help - General Term for a Sequence

  1. #1
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    General Term for a Sequence

    Find an of {an} given a1 = 1, and an+1 = Sn + 2n + 1
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  2. #2
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    Hello, MATNTRNG!


    I hope the S is a typo . . .


    \text{Find }a_n\text{ of }\{a_n\}.\;\text{ given: }\:a_1 = 1,\;\;a_{n+1} = a_n + 2n + 1

    Crank out the first few terms: . 1,\;6,\;13,\;22,\;33,\;46,\;61,\;\hdots


    Take the difference of consecutive terms,
    . . then the differences of the differences, and so on.

    . . \begin{array}{cccccccccccc}<br />
\text{Sequence:} & 1 && 6 && 13 && 22 && 33 && 46 \\<br />
\text{1stn diff:} && 5 && 7 && 9 && 11 && 13 \\<br />
\text{2nd diff:} &&& 2 && 2 && 2 && 2 && 2 \end{array}


    Since the second differences are constant,
    . . the generating function is of the second degree, a quadratic.

    The general quadratic function is: . f(n) \:=\:an^2 + bn + c


    Use the first three values of the function and substitute:

    . . \begin{array}{ccccc}<br />
f(3) = 13: & 9a + 3b + c &=& 13 & [1] \\<br />
f(2) = 6: & 4a + 2b + c &=& 6 & [2] \\<br />
f(1) = 1: & a + b + c & = & 1 & [3] \end{array}

    . . \begin{array}{ccccc}<br />
\text{Subtract [1] - [2]:} & 5a + b &=& 7 & [4] \\<br />
\text{Subtract [2] - [3]:} & 3a + b &=& 5 & [5] \end{array}

    . . \text{Subtract [4] - [5]:} \;\;2a \:=\:2 \quad\Rightarrow\quad \boxed{a \:=\:1}

    Substitute into [5]: . 3(1) + b \:=\:5 \quad\Rightarrow\quad\boxed{ b \:=\:2}

    Substitute into [3]: . 1 + 2 + c \:=\:1 \quad\Rightarrow\quad\boxed{ c \:=\:-2}


    Therefore, the generating function is: . f(n) \;=\;n^2 + 2n - 2

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