Prove that the sum of the Lagrange (interpolation) coefficients is equal to 1 Prove that the sum of the Lagrange (interpolation) coefficients is equal to 1.
Please suggest me a book-reference or give a solution for me. Thanks a lot in advance. 
If $f = \sum_{i=0}^nf(x_i)L_i(x)$ then one has to prove  $\sum_{i=0}^nL_i(x)=1$ where
 $L_i(x)=\frac{(x-x_0)...(x-x_{i-1})(x-x_{i+1})...(x-x_n)}{(x_i-x_0)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_n)}$,   $i=0 \cdots n$
 A: Let $p$ be the unique polynomial of degree at most $n-1$ that passes through the points $(x_1,1),...,(x_n,1)$. Then $p(x) = \sum_{i=1}^nL_i(x)$.
The polynomial $q(x) = 1$ has degree zero and passes through these points.
Aside: The key fact is that if $r$ is a polynomial of degree at most $n$ and
$r(x_k) = 0$ at $n+1$ distinct points $x_k$, then $r= 0$.
A: Let me propose one more method of proof.
"Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation", describes a determinant form of Lagrange polynomial that interpolates $(a_0;b_0)$, $\dots$, $(a_n;b_n)$
$$
P(x) = (-1)
\frac{
    \det
    \begin{pmatrix}
        0       & b_0       & b_1       & \cdots & b_n       \\
        x^n     & a_0^n     & a_1^n     & \cdots & a_n^n     \\
        x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots  & \cdots    & \cdots    & \cdots & \cdots    \\
        1       & 1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        a_0^n     & a_1^n     & \cdots & a_n^n     \\
        a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}.
$$
Obviously, setting all $b_i = 1$ we'll get the sum of Lagrange polynomials $L_i$, because
$$
    P(x) = \sum_{i=0}^n\, \underbrace{b_i}_{=1} L_i(x) = \sum_{i=0}^n L_i(x).
$$
Now just simplify everything
$$
P(x) = (-1)
\frac{
    \det
    \begin{pmatrix}
        0       & 1         & 1         & \cdots &   1       \\
        x^n     & a_0^n     & a_1^n     & \cdots & a_n^n     \\
        x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots  & \cdots    & \cdots    & \cdots & \cdots    \\
        1       & 1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        a_0^n     & a_1^n     & \cdots & a_n^n     \\
        a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
} =
(-1)
\frac{
    \det
    \begin{pmatrix}
        -1      & 0         & 0         & \cdots &   0       \\
        x^n     & a_0^n     & a_1^n     & \cdots & a_n^n     \\
        x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots  & \cdots    & \cdots    & \cdots & \cdots    \\
        1       & 1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        a_0^n     & a_1^n     & \cdots & a_n^n     \\
        a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}.
$$
Laplace expansion along the first row leads to
$$
P(x) =
\frac{
    \det
    \begin{pmatrix}
        a_0^n     & a_1^n     & \cdots & a_n^n     \\
        a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        a_0^n     & a_1^n     & \cdots & a_n^n     \\
        a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
} = 1.
$$
For practical example you may want to check "Workbook on mapping simplexes affinely", section "Lagrange interpolation".
NOTE. Using Laplace expansion along the first column in the numerator you can get expressions for coefficients at $x^i$, thus determinant form of Lagrange interpolation can be quite convenient in some cases.
