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

Banguizb 2022-07-07 Answered
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 > 0         a n d       a 1 = 1 ,         a 2 = 4
This 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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Answers (1)

isscacabby17
Answered 2022-07-08 Author has 13 answers
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 + 2
this 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 2
That is
a n + 2 a n + 1 = a 2 a 1 + 2 n + 2 n + 1 2 = 2 n + 2 n + 1 + 1
By 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 )

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