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You should notice that adding the first and the second terms yields the third term, then adding the second and the third term yields the fourth term, hence, using inductive method, you should prove the following formula, such that:
`x_(n+2) = x_(n+1) + x_(n) => x_(n) = x_(n+2) - x_(n+1) `
You need to perform th steps of mathematical induction, hence, you need to start by verifying that` x_1 = x_3 - x_2` such that:
`x_1 = 1, x_2 = 4, x_3 = 5`
`1 = 5 - 4 => 1 = 1` valid
You need to perform the next step, hence, assuming that P(k) is valid, then `P(k+1)` is alos valid.
`P(k): x_k = x_(k+2) - x_(k+1)` valid
You need to prove that `P_(k+1): x_(k+1) = x_(k+3) - x_(k+2)` is also valid.
`x_(k+1) = x_(k+3) - x_(k+2)`
Since `x_(k+2) = x_k + x_(k+1)` is valid yields:
`x_(k+1) = x_(k+3) - x_k - x_(k+1) => 2x_(k+1) = x_(k+3) - x_k`
`x_(k+1) = (x_(k+3) - x_k)/2` valid
Hence, evaluating the reccurence formula of the given sequence, yields `x_(n) = x_(n+2) - x_(n+1).`
The sequence that has to be generated by the recursive formula has the terms 1, 4, 5, 9, 14, ...
It can be seen from the terms provided that starting with the third term, each term is the sum of the preceding 2 terms.
`T_1 = 1` and `T_2 = 4` .
Starting with `T_3` , `T_n = T_(n-1) + T_(n-2)`
The recursive formula that generates the sequence 1, 4, 5, 9, 14, ... is `T_n = T_(n-1) + T_(n-2)` with `n>= 3` .
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