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Math Help - Sequences

  1. #1
    Member GAdams's Avatar
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    Sequences

    Hi

    Could somebody explain how I would go about doing the attached problem. I don't really get it at all so a step-by-step thing would be good.

    Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    Hi

    Could somebody explain how I would go about doing the attached problem. I don't really get it at all so a step-by-step thing would be good.

    Thanks.
    In this notation, the nth term of the sequence is identified by the subscript attached to it. So, the first term is a_1, the second is a_2, etc. you have to find a_2 through a_5. here, we use the recursive formula given. the formula says:

    a_1 = 1, \mbox { } a_{n + 1} = a_n^2 - 1

    we have to replace the n's in the formula with the appropriate n and evaluate.

    so to find a_2, we notice that that is a_{1 + 1}, so we can take n = 1 to find a_2

    plugging in n=1 into the formula, we find that:

    a_2 = a_1^2 - 1

    now evaluate that. the rest are similar
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  3. #3
    Member GAdams's Avatar
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    I get:

    1, 0, -1, 0, -1
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    I get:

    1, 0, -1, 0, -1
    correct
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  5. #5
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    Hello, GAdams!

    Give the first five terms.

    . . a_1 = 1\qquad a_{n+1} \:=\:a_n^2-1\qquad(n \,=\,1,2,3,\cdots)
    You have to read what it says . . .


    a_1\,=\,1
    . . It says: the first term is 1.

    a_{n+1} \:=\:a_n^2-1
    . . It says: Each term is the square of the preceding term, minus 1.


    So we have:

    \begin{array}{ccccccccc}a_1 & = & 1 \\<br /> <br />
a_2 & = & a_1^2 - 1  & = & 1^2-1 & \Rightarrow & a_2 & = & 0\\<br /> <br />
a_3 & = & a_2^2-1 & = & 0^2-1 & \Rightarrow & a_3 & = & \text{-}1 \\<br /> <br />
a_4 & = & a_3^2-1 & = & (\text{-}1)^2-1 & \Rightarrow & a_4 & = & 0 \\<br /> <br />
a_5 & = & a_4^2-1 & = & 0^2-1 & \Rightarrow & a_5 & = & \text{-}1<br />
\end{array}

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