Answer
Yes!
The proof relies on the composition of age-groups within the population of each month. Let R(M) represent the sequence called R, where the rabbits live for M months. Then, let
An represent the # of newborns in the nth month, and
Bn represent the # of 1-month olds in the nth month, and
Cn represent the # of 2-month olds in the nth month, and
..., and
Zn represent the # of M-month olds in the nth month.
Then, because the number of mature, adult rabbits in the previous month determines the number of newborns in the current month, we say that
Also, because the rabbits age as time goes on, we say that
Due to the length of the proof, it will not appear in this post. Although, I will explain how the proof works out. So, the above equations allow me to say this
because the population in the nth month is only comprised of the M age-groups.
Then, I can expand the An term and substitute for Bn through Zn to give us
We do this expansion again, to give us two of the A terms. This is very important, because in substitution, the A term is the only term which gives us all terms B through Z. Then, once we have the two A terms, we can group A through Z (with the same subscript, say n-2) as R(M)n-2. Continue to substitute and regroup until all the terms (A through Z) have been regrouped into populations (R(M)). The final result will be
This is exactly what I wanted to confirm.
The proof relies on the composition of age-groups within the population of each month. Let R(M) represent the sequence called R, where the rabbits live for M months. Then, let
An represent the # of newborns in the nth month, and
Bn represent the # of 1-month olds in the nth month, and
Cn represent the # of 2-month olds in the nth month, and
..., and
Zn represent the # of M-month olds in the nth month.
Then, because the number of mature, adult rabbits in the previous month determines the number of newborns in the current month, we say that
Also, because the rabbits age as time goes on, we say that
Due to the length of the proof, it will not appear in this post. Although, I will explain how the proof works out. So, the above equations allow me to say this
because the population in the nth month is only comprised of the M age-groups.
Then, I can expand the An term and substitute for Bn through Zn to give us
We do this expansion again, to give us two of the A terms. This is very important, because in substitution, the A term is the only term which gives us all terms B through Z. Then, once we have the two A terms, we can group A through Z (with the same subscript, say n-2) as R(M)n-2. Continue to substitute and regroup until all the terms (A through Z) have been regrouped into populations (R(M)). The final result will be
This is exactly what I wanted to confirm.
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