<s>[INST] Is it possible to have a rule which generates: 2, 4, 6, 8, 10, 12, 14, 16, -23? This is on Lagrange Interpolations . . .
Is it possible to have a rule which generates the sequence: 2, 4, 6, 8, 10, 12, 14, 16, -23?
The hint that he gave us is to use Summation Products, the only thing I can come up with is $\sum_{i=1}^{8}{2}$.  I cannot come up with any formula for a Summation Product that would produce something like a 2 + 2 + 2 + . . .
I know how to use the equation for Lagrange Interpolation, but I cannot see how to apply it here . . .
 [/INST] Thomas Andrews's solution is the best one, but you may also consider this simplistic but longer and bulky solution:
$$
\begin{array}{lccr} \\
 f(x) & = &   &  a_1\times(-1)^{1-1}\times\frac{1}{9!}(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
      &   & + &  a_2\times(-1)^{2-1}\times\frac{2}{9!}(x-1)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
      &   & + &  a_3\times(-1)^{3-1}\times\frac{3}{9!}(x-1)(x-2)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
      &   & + &  a_4\times(-1)^{4-1}\times\frac{4}{9!}(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)(x-8)(x-9) \\
      &   & + & a_5\times(-1)^{5-1}\times\frac{5}{9!}(x-1)(x-2)(x-3)(x-4)(x-6)(x-7)(x-8)(x-9) \\
      &   & + & a_6\times(-1)^{6-1}\times\frac{6}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)(x-8)(x-9) \\
      &   & + & a_7\times(-1)^{7-1}\times\frac{7}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-8)(x-9) \\
      &   & + & a_8\times(-1)^{8-1}\times\frac{8}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-9) \\
      &   & + & a_9\times(-1)^{9-1}\times\frac{9}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8) \\
\end{array} $$
Where $ a_1 = 2 $, $ a_2 = 4 $, $ a_3 = 6 $, $ a_4 = 8 $, $ a_5 = 10 $, $ a_6 = 12 $, $ a_7 = 14 $, $ a_8 = 16 $ and $ a_9 = -23$.
This may look very long if you are working analytically. But if you are implementing numerical methods in computer, this can be automated and greatly simplified by a simple software algorithm as a general solution to generate arbitrary discrete sequences.
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