Fibonacci hare problem?

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Jovian hares are hermaphrodite. Each mature Jovian hare produces one leveret during each breeding cycle. Each leveret takes two breeding cycles to mature into a fully-grown hare, and then lives for ever. Starting with a single newly-born Jovian leveret,how many hares (including leverets) are there after 47 breeding cycles?

This is the question that I am being asked to consider. I have found the equation to calculate the number of hares and leverets after $n$ breeding cycles. It is $M_{n}=M_{n-1}+M_{n−3}$ and have found the number of hares and leverets that will present after 47 breeding cycles. How would I go about proving that $M_{n}=M_{n-1}+M_{n−3}$? By induction or what?

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You don't need induction, let $M_n$ be the number of hares that exist at time $n\geq 3$.

Then the number of leverets at time $M_n$ is equal to the number of leverets that where already present at time $n-1$ plus the number of leverets that are born in time $n$. The first number is $M_{n-1}$, how many leverets are born in time $n$? This is equal to the number of adult hares that where present at time $n-1$, and this is clearly equal to the number of hares that where present at time $n-3$.

Therefore we have $M_0=1,M_1=1,M_2=1,M_3=2$ and we can calculate the remaining terms with the recursion. In particular you need $M_{47}$