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$\newcommand{\Fib}{\operatorname{Fib}}$I am trying to reduce this expression for the $n$th term of sequence $G$.

$G[n]=\Fib(4) \times \Fib(n-1) + \Fib(5) \times \Fib(n-2) + \Fib(6) \times \Fib(n-3)+ \cdots +\Fib(n+3) \times \Fib(1)$

Here $\Fib()$ is the Fibonacci sequence $1,2,3,5,8,13, \ldots$

How can I simplify expression for $G[n]$?

I have read the various identities given on Wikipedia but have not been able to apply them. Any help will be highly appreciated, as I have been stuck on this for two days.

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    possible duplicate of [Summation of series of product of Fibonacci numbers](http://math.stackexchange.com/questions/190586/summation-of-series-of-product-of-fibonacci-numbers)2012-09-03
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    See [Robert Israel’s answer](http://math.stackexchange.com/a/190597/12042) to the question cited above.2012-09-03
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    @BrianM.Scott:Sir I don't consider it as an exact duplicate .Here I would like to add that I need a formula that doesn't require any floating point computation.2012-09-03
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    If you don't want to use floating point, you are stuck with a recursion formula you have to evaluate step by step after manually computing $G(1)), G(2), G(3), G(4)$: $G(n) = 2G(n-1)+G(n-2)-2G(n-3)-G(n-4)$2012-09-03

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