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    2a_n +a_{n-1} =0

    where a_0 =2

    i get as the solution a_n =2(-2)^n which should be -½ not -2

    but what is a_1 and\ a_2 because that's what i kinda guess my answer off i got -4 and 8 which of course is wrong so can someone point me in the right direction?
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    (Original post by will'o'wisp2)
    2a_n +a_{n-1} =0

    where a_0 =2

    i get as the solution a_n =2(-2)^n which should be -½ not -2

    but what is a_1 and\ a_2 because that's what i kinda guess my answer off i got -4 and 8 which of course is wrong so can someone point me in the right direction?
    You can rewrite your relation as a_{n+1}=-\frac{1}{2}a_n by mapping n \mapsto n+1 (ie increasing each subscript by 1) and rearranging for a_{n+1}

    So a_1=-\frac{1}{2}(2)=-1 and a_2=\frac{1}{2} - two conditions which are not satisfied by your solution.
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    (Original post by RDKGames)
    You can rewrite your relation as a_{n+1}=-\frac{1}{2}a_n by mapping n \mapsto n+1 (ie increasing each subscript by 1) and rearranging for a_{n+1}

    So a_1=-\frac{1}{2}(2)=-1 and a_2=\frac{1}{2} - two conditions which are not satisfied by your solution.
    ah coolio, that's another thing then i need to note down

    thanks a bunch
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    (Original post by will'o'wisp2)
    ah coolio, that's another thing then i need to note down

    thanks a bunch
    No problem, also I meant to say that a_1=2 then, and a_2=-1 with a_3=\frac{1}{2}. Then proceed to find your solution, but instead of a_n write a_{n+1} as we increased them by 1. Then reduce the subscripts by 1 to their original state after you achieve it, as then the context of the question and conditions are satisfied.

    You can skip all of this increasing/decreasing if you just let k=n-1 to begin with in your Q and work in k's before you convert back to n's
 
 
 
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