# proof of $\sum\nolimits_{i = 1}^{n } {\prod\nolimits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i – x_j }}} } = 1$

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• Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$

#### Solutions Collecting From Web of "proof of $\sum\nolimits_{i = 1}^{n } {\prod\nolimits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i – x_j }}} } = 1$"

Consider the Lagrange interpolation of the polynomial $f(x)=x^{n-1}$ with interpolation points $x_1,\ldots,x_n$. We have:
$$x^{n-1}=\sum_{i=1}^{n}x_i^{n-1}\prod_{j\neq i}\frac{x-x_j}{x_i-x_j},$$
so, by comparing the leading coefficients of RHS and LHS, the result follows.

For $n>1$, the partial fractional decomposition of
$$\frac1z\prod_{j=1}^{n-1}\frac{z}{z-x_j} =\frac{z^{n-2}}{\prod\limits_{j=1}^{n-1}(z-x_j)} =\sum_{i=1}^{n-1}\frac{A_i}{z-x_i}\tag{1}$$
using the Heaviside method yields
$$A_i=\frac{x_i^{n-2}}{\prod\limits_{\substack{j=1\\j\ne i}}^{n-1}(x_i-x_j)} =\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{2}$$
Combine $(1)$ and $(2)$:
\begin{align} \prod_{j=1}^{n-1}\frac{z}{z-x_j} &=\sum_{i=1}^{n-1}A_i\frac{z}{z-x_i}\\ &=\sum_{i=1}^{n-1}\frac{z}{z-x_i}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\\ &=\sum_{i=1}^{n-1}\left(1-\frac{x_i}{x_i-z}\right)\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{3} \end{align}
Set $z=x_n$ in $(3)$:
$$\prod_{j=1}^{n-1}\frac{x_n}{x_n-x_j} =\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j} -\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}\tag{4}$$
Add the second term of the right side of $(4)$ to both sides:
$$\sum_{i=1}^n\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j} =\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{5}$$
Noting that the case $n=1$ follows vacuously, using $(5)$ and induction proves that
$$\sum_{i=1}^n\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}=1\tag{6}$$