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Math Help - Formula for nth term of sequence

  1. #1
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    Formula for nth term of sequence

    Suppose that a_{1} = 3 and a_{n+1}=a_{n}\left(a_{n}+2\right). Find a formula for the nth term of this sequence.
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  2. #2
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    Hint

    Here's a hint: consider the fact that (2^k-1)[(2^k-1)+2]=(2^k-1)(2^k+1)=2^{2k}-1.

    --Kevin C.
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  3. #3
    Senior Member Shanks's Avatar
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    a_{n+1}=a_{n}\left(a_{n}+2\right)\text{ implies }a_{n+1}+1=a_{n}\left(a_{n}+2\right)+1,\text{ that is }a_{n+1}+1=\left(a_{n}+1\right)^2
    Construct a new sequence by letting b_n=a_n+1, you can take it from here.
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  4. #4
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    Hello, Pn0yS0ld13r!

    I found the answer by inspection . . .


    Suppose that a_1 = 3 and a_{n+1}\:=\:a_n\left(a_n+2\right).

    Find a formula for the n^{th} term of this sequence.
    I cranked out the first few terms . . .

    . . \begin{array}{|c|ccc|}<br />
n & & a_n & \\ \hline<br />
1 & 3 &=& 2^2-1 \\<br />
2 & 15 &=& 2^4-1\\<br />
3 & 255 &=& 2^8-1\\<br />
4 & 65,\!535 &=& 2^{16}-1 \\<br />
5 & 4,\!294,\!967,\!295 &=& 2^{32}-1 \\<br />
\vdots & \vdots & & \vdots<br />
\end{array}


    And saw that: . a_n \;=\;2^{2^n}-1

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