Let f(x) be a polynomial of degree 2n for some natural number n such that f(x)=1/x for x=1,2,3,...2n+1. What is the value of f(2n+3)?
Start with defining a series of polynomials g_k(x).
For each integer k in the range [1,2n+1] let g_k(x) equal prod {i=1 to 2n+1 except i=k} (x-i)/(k-i).
g_k(x) will have zeros at 2n of the integers in [1,2n+1] except for x=k where g_k(k) will equal 1.
Then f(x) = sum {k=1 to 2n+1} g_k(x)*(1/k).
Each of the 2n+1 terms of the summation for f(x) looks like
(x-1)*(x-2)*...*(x-k+2)*(x-k+1) * 1 * (x-k-1)*(x-k-2)*...*(x-2n)*(x-2n-1)
-------------------------------------------------------------------------
(k-1)*(k-2)*...* (2) * (1) * k * (-1) * (-2) *...*(k-2n)*(k-2n-1)
Each numerator equals (x-1)!/((x-2n-2)!*(x-k)) when x is unequal to k, which is fine since we are to evaluate at x=2n+3.
Substituting x=2n+3 simplifies the numerators to (2n+2)!/(2n+3-k)
Each denominator is equal to (-1)^(k-1)*k!*(2n+1-k)!.
Then f(2n+3) = sum {k=1 to 2n+1} (2n+2)! / ((-1)^(k-1)*(2n+3-k)*k!*(2n+1-k)!)
This can be simplified a bit to sum {k=1 to 2n+1} (-1)^(k-1) * (2n+2)/(2n+3-k) * C(2n+1,k)
This is as far as I got.
Thoughts
