Please help with this interesting sum proof Hi, I am trying to prove the following identity. I obviously can't see something here. $\displaystyle \sum_{1}^{n}\frac{\prod_{r=1}^{n1}(y_kz_r)}{\prod_{1<=r<=n, r\ne k}(y_ky_r)}=1 $ Where $\displaystyle y_1,y_2,...y_n$ are different numbers and $\displaystyle z_1,z_2,...z_{n1}$ is set of any numbers. I obviously can't see something. Any hints? 
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$\displaystyle {\Large\sum_{k=1}^{n}}\frac{\displaystyle {\small \prod_{\large r=1}^{\large n1}}(y_kz_r)}{\displaystyle {\small \prod_{\large1\leqslant r\leqslant n, r\ne k}}(y_ky_r)}=1 $ 
It has the look of Lagrange interpolation. 
I finally did it Hi, I think I did it. Actually, least common multiple of all sum members is $\displaystyle \prod_{i=1}^{n1}\prod_{k=i+1}^{n}(y_iy_k)$. In numerator, we have the sum with member with ordinal number i is $\displaystyle (1)^{i+1} \prod_{k=1}^{n1}(y_iz_k)\prod_{1\le k\le n1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_ky_r)$ For shorter notation, let's have $\displaystyle \pi(i) = \prod_{1\le k\le n1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_ky_r)$ When we develop the the product, we get $\displaystyle (y_i^{n1} + \sum_{k=1}^{n1}((1)^{k}y_i^{k1}\sum_{r=1}^{\binom{n1}{k}}C(k,r)))\cdot \pi(i)$ where C(k,r) is rth combination of k elements on array $\displaystyle (z_1,z_2,...,z_{n1})$ When we get back to the sum in the numerator, we get $\displaystyle \sum_{i=1}^{n}y_i^{n1}\cdot \pi(i) + \sum_{k=1}^{n1}(1)^{k}\sum_{r=1}^{\binom{n1}{k}}C(k,r)\cdot \sum_{i=1}^{n}y_i^{k1}\cdot \pi(i)$ As we now got multipliers of zarray (their combinations) we have to prove that every $\displaystyle \sum_{i=1}^{n}y_i^{r}\cdot \pi(i)= \sum_{i=1}^{n}y_i^{r}\prod_{1\le k\le n1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_ky_r)=0$ for r < n1. Approach used here uses the fact that all elements have the same polynomial degree and greatest degree of particular variable is n2. Thus for each element (possible repeatable combination of yarray) we will have its negative pair as 1 is equally combined as all yarray elements. Therefore result has to be zero. When we look $\displaystyle \sum_{i=1}^{n}y_i^{n1}\cdot \pi(i)$ We see that it can't be zero as $\displaystyle y_i^{n1}$ can't be found in other sum elements where as maximal degree of particular element there is n2. However we can look for "similar" one that has $\displaystyle y_l^{n1}y_y{n2}$ and the same products in continuation. We then $\displaystyle y_l^{n2}y_i^{n2}(y_iy_l)(rest \; of \; product)$ We now halved number of sum elements. This can be repeated recursively n1 times and then we get one product that is $\displaystyle \prod_{i=1}^{n1}\prod_{k=i+1}^{n}(y_iy_k)$ which gives what is wanted. 
I finally did it For some reason my yesterday reply is not published. I didn't actually finished the proof as I illustrated idea behind the proof. Waiting for moderator to publish it so I can finalize it. 
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And now to get to the point Please first note that I made mistake by not including $\displaystyle (1)^{k}$ and $\displaystyle (1)^{i+1}$ in sums with $\displaystyle y_i$ elements which you will see below. In divisor from the beginning we have $\displaystyle \prod_{i=1}^{n1}\prod_{k=i+1}^{n}(y_iy_k) = (1)^{n1}det(V_{n1})$ Where V is Vandermonde matrix of order of n1 (n x n dimensions). $\displaystyle \pi(i) = (1)^{n2}det(V_{n2}(i))$ where V(i) is Vandermonde matrix of order of n2 where $\displaystyle y_i$ is missing. Then the sum $\displaystyle \sum_{i=1}^{n}(1)^{i}y_{i}^{k1}\cdot \pi(i)$ is determinant of matrix which has $\displaystyle (1)^{n2}V(i)$ as cofactor and vector $\displaystyle (y_1^{k1}, ...,y_n^{k1})$ where $\displaystyle 1 \le k \le n1$ so it is less or equal to n2. This means that in V(i) there is already such column of values $\displaystyle \begin{bmatrix} 1& y_1& \cdots & y_1^{k1}& y_1^{k1}& \cdots& y_1^{n2}&\\ \vdots & \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \\ 1& y_n& \cdots& y_n^{k1}& y_n^{k1}& \cdots& y_n^{n2}& \end{bmatrix} $ And determinant of each such matrix is zero. In $\displaystyle \sum_{i=1}^{n}(1)^{i+1}y_i^{n1}\cdot \pi(i)$ we have similar situation with V(i) as cofactors. However we now have vector $\displaystyle (y_1^{n1}, ...,y_n^{n1})$ as matrix column so we are getting Vandermonde matrix of order n1. Therefore we have $\displaystyle \sum_{i=1}^{n}(1)^{i+1}y_i^{n1}\cdot \pi(i)=(1)^{n1}det(V)$ At the end $\displaystyle \frac{(1)^{n1}det(V)}{(1)^{n1}det(V)}=1$ 
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