# Problem 101: Optimum polynomial
If we are presented with the first k terms of a sequence it is
impossible to say with certainty the value of the next term, as there
are infinitely many polynomial functions that can model the sequence. As
an example, let us consider the sequence of cube numbers. This is
defined by the generating function, un = n3: 1, 8, 27, 64, 125, 216, ...
Suppose we were only given the first two terms of this sequence. Working
on the principle that "simple is best" we should assume a linear
relationship and predict the next term to be 15 (common difference 7).
Even if we were presented with the first three terms, by the same
principle of simplicity, a quadratic relationship should be assumed. We
shall define OP(k, n) to be the nth term of the optimum polynomial
generating function for the first k terms of a sequence. It should be
clear that OP(k, n) will accurately generate the terms of the sequence
for n ≤ k, and potentially the first incorrect term (FIT) will be OP(k,
k+1); in which case we shall call it a bad OP (BOP). As a basis, if we
were only given the first term of sequence, it would be most sensible to
assume constancy; that is, for n ≥ 2, OP(1, n) = u1. Hence we obtain the
following OPs for the cubic sequence: OP(1, n) = 1 1, 1, 1, 1, ... OP(2,
n) = 7n−6 1, 8, 15, ... OP(3, n) = 6n2−11n+6      1, 8, 27, 58, ...
OP(4, n) = n3 1, 8, 27, 64, 125, ... Clearly no BOPs exist for k ≥ 4. By
considering the sum of FITs generated by the BOPs (indicated in red
above), we obtain 1 + 15 + 58 = 74. Consider the following tenth degree
polynomial generating function: un = 1 − n + n2 − n3 + n4 − n5 + n6 − n7
+ n8 − n9 + n10 Find the sum of FITs for the BOPs.