I would like to find the closed form of the sequence given by a n + 2 = 2 a n + 1 − a n + 2 n + 2 , n &gt; 0 a n d a 1 = 1 , a 2 = 4This task is in the topic of differential and difference equation. I don't know how to start solving this problem and what are we looking for? ( a n , a n + 2 )I do know how to solve the following form a n + 2 = 2 a n + 1 − a n using linear algebra as well. The actual problem I encountered the obstructionist term 2 n + 2 Are there some kind of variational constant method for recursive linear sequences,?I only now this method for linear ODE with constant coefficient.But I believe that such method could be doable here as well. Can any one provide me with a helpful hint or answer?.

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I would like to find the closed form of the sequence given by      a          n      +      2        =  2      a          n      +      1        −      a    n    +      2    n    +  2  ,              n  &amp;gt;  0              a  n  d               a    1    =  1  ,                  a    2    =  4This task is in the topic of differential and difference equation. I don&#039;t know how to start solving this problem and what are we looking for? (      a    n    ,      a          n      +      2      )I do know how to solve the following form      a          n      +      2        =  2      a          n      +      1        −      a    n  using linear algebra as well. The actual problem I encountered the obstructionist term             2      n        +    2  Are there some kind of variational constant method for recursive linear sequences,?I only now this method for linear ODE with constant coefficient.But I believe that such method could be doable here as well. Can any one provide me with a helpful hint or answer?.

isscacabby17 · Accepted Answer

Answer: By telescoping twice we obtain the following formula            a      n        =          2              n              +    n    (    n    −    2    )  Enforcing       X    n    =      a          n      +      1        −      a    n   yields that      a          n      +      2        =  2      a          n      +      1        −      a    n    +      2    n    +  2    ⟹    (      a          n      +      2        −      a          n      +      1        )  −  (      a          n      +      1        −      a    n    )  =      2    n    +  2this leads to            X              n        +        1              −          X      n        =          2      n        +    2  By telescopic sum we have      X          n      +      1        −      X    1    =      ∑          k      =      1              n            X          k      +      1        −      X    k    =      ∑          k      =      1              n        [      2    k    +  2  ]  =  2  n  +      2          n      +      1        −  2That is      a          n      +      2        −      a          n      +      1        =      a          2        −      a    1    +  2  n  +      2          n      +      1        −  2  =  2  n  +      2          n      +      1        +  1By telescopic once more we remain with      a          n      +      2        =      a    2    +      ∑          k      =      1              n            [    2    k    +          2              k        +        1              +    1    ]    =  4  +  n  (  n  +  1  )  +      2          n      +      2        −  4  +  n  =      2          n      +      2        +  n  (  n  +  2  )Finally            a      n        =          2              n              +    n    (    n    −    2    )

I would like to find the closed form of the sequence given by a <mrow class="MJX-TeX

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