Discrete Mathematics- An Open Introduction - 3rd Edition, 2016a

164 2. Sequences
3. 1, 1, 2, 3, 5, 8, 13, . . .
Solution.
1. As we saw in Example 2.3.1, this sequence is not ∆ k -constant
for any k. Therefore the closed formula for the sequence is
not a polynomial. In fact, we know the closed formula is
a n 2 n , which grows faster than any polynomial (so is not a
polynomial).
2. The sequence of first differences is 7, 43, 133, 301, 571, . . .. The
second differences are: 36, 90, 168, 270, . . .. Third difference:
54, 78, 102, . . .. Fourth differences: 24, 24, . . .. As far as we can
tell, this sequence of differences is constant so the sequence
is ∆ 4 -constant and as such the closed formula is a degree 4
polynomial.
3. This is the Fibonacci sequence. The sequence of first differences
is 0, 1, 1, 2, 3, 5, 8, . . ., the second differences are 1, 0, 1, 1, 2, 3, 5 . . ..
We notice that after the first few terms, we get the original
sequence back. So there will never be constant differences,
so the closed formula for the Fibonacci sequence is not a
polynomial.
Exercises
1. Use polynomial fitting to find the formula for the nth term of the
sequence (a n ) n≥0 which starts,
0, 2, 6, 12, 20, . . . .
2. Use polynomial fitting to find the formula for the nth term of the
sequence (a n ) n≥0 which starts,
1, 2, 4, 8, 15, 26 . . . .
3. Use polynomial fitting to find the formula for the nth term of the
sequence (a n ) n≥0 which starts,
2, 5, 11, 21, 36, . . . .
4. Use polynomial fitting to find the formula for the nth term of the
sequence (a n ) n≥0 which starts,
3, 6, 12, 22, 37, 58, . . . .