The sum of Cotes coefficients

Question

Newton-Cotes coefficients is as follows $$ C_{i}^{(n)}=\frac{(-1)^{n-i}}{i !(n-i) ! n} \int_{0}^{n} \prod_{\substack{j=0 \\ j \neq i}}^{n}(t-j) d t $$ How can we prove that $$ \sum_{i=0}^nC_i^{(n)}=1 $$

Answer 1

If you consider the polynomial

$$ F_i(t) =\frac{(-1)^{n-i}}{i !(n-i) ! n} \prod_{\substack{j=0 \\ j \neq i}}^{n}(t-j) d t $$

And look at the values for $t = 0,1,2,\dots, n$. It is only non-zero at $t = i$, and the value equals $\frac{1}{n}$.

Therefore

$\sum_{i=0}^n F_i(t)$ is a polynomial of degree $n$, and equals $\frac{1}{n}$ at all integer nodes $t=0,1,2,\dots, n$, thus it must be constant $\frac{1}{n}$. Its integral equals one.