Square of a Sequence of Numbers

Question

This is a simple question, but I could not find the solution.

What is the compact form of expansion of $(n_1+n_2+\ldots+n_k)^2$

Is it: $\sum_{i=1}^k n_i^2 + 2\sum_{i=1}^{n-1} \sum_{j=i+1}^{n}n_in_j$

Is there any reference?

Answer 1

By using distributivity and commutativity we have \begin{align*} \left( \sum_{i=1}^k n_i \right)^2 &= \left( \sum_{i=1}^k n_i \right) \cdot \left( \sum_{j=1}^k n_j \right) = \sum_{i=1}^k \left(n_i \cdot \sum_{j=1}^k n_j \right) = \sum_{i=1}^k \sum_{j=1}^k \left(n_i \cdot n_j \right) \\ &= \sum_{i,j=1}^k n_i n_j = \sum_{1 \leq i = j \leq k} n_i n_j + \sum_{1 \leq i < j \leq k} n_i n_j + \sum_{1 \leq j < i \leq k} n_i n_j \\ &= \sum_{i=1}^k n_i^2 + \sum_{1 \leq i < j \leq k} n_i n_j + \sum_{1 \leq i < j \leq k} \underbrace{n_j n_i}_{= n_i n_j} = \sum_{i=1}^k n_i^2 + 2 \sum_{1 \leq i < j \leq k} n_i n_j \\ &= \sum_{i=1}^k n_i^2 + 2 \sum_{i=1}^{k-1} \sum_{j=i+1}^k n_i n_j. \end{align*}

Answer 2

It is right in a commutative ring.

Just compute it like this way: $(n_1+n_2+...)(n_1+n_2+...)=\Sigma_{i=j}n_i n_j+\Sigma_{i<j}n_i n_j+\Sigma_{i>j}n_i n_j=\Sigma_{i=j}n_i n_j+2\Sigma_{i<j}n_i n_j$