Do sequences with an equivalent second difference have a quadratic nth term?
Yes they do.
When the 1st differences are constant, then we have an arithmetic sequence, with the nth term given by a linear function.
When it is the 2nd differences that are constsant, then the nth term is given by a quadratic function. See below.
Compound Arithmetic Sequence
b0 b1 b2 b3 b4 b5
c0 c1 c2 c3 c4 -------------- 1st differences
d d d d ------------------- 2nd differences = constant
We have here an irregular sequence (b_n) = (b_1, b_2, b_3, ..., b_k, ...).
The (1st) differences between the elements of (b_n), the 1st differences, are non-constant.
However, the (2nd) differences between the elements of (c_n) are constant. This makes (c_n) a regular arithmetic sequence, and so we can write,
cn = c0 + nd, n = 0,1,2,3,...
The sequence (b_n) is non-regular, but we can write,
b_(n+1) = b_n + c_n
Using the expression for the c-sequence,
b_(n+1) = b_n + c_0 + nd
Solve the recurrence relation for b_n
Develop the terms of the sequence.
b1 = b0 + c0
b2 = b1 + c0 + 1.d = b0 + 2.c0 + 1.d
b3 = b2 + c0 + 2.d = b0 + 3.c0 + (1 + 2).d
b4 = b3 + c0 + 3.d = b0 + 4.c0 + (1 + 2 + 3).d
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b(n+1) = b_0 + (n+1).c0 + sum[k=1..n](k) * d
b_(n+1) = b_0 + (n+1).c0 + (nd/2)(n + 1)
b_(n+1) = b_0 + (n+1)(c_0 + nd/2)
We could also write c0 = b1 – b0 in the above expression.