Reducing an algebraic summation expression

Question

Consider the following elements of an algebraic series $$\sum_{i=1}^Nc(\theta_i)$$ $$\frac{1}{N}\sum_{j=1}^N\sum_{i=1,i\neq j}^Nc(\theta_i)$$ $$\frac{1}{N-1}\frac{1}{N}\sum_{k=1}^N\sum_{\array{j=1 \\j\neq k}}^N\sum_{\array{i=1\\i\neq j\\i\neq k}}^Nc(\theta_i)$$ $$\frac{1}{N-2}\frac{1}{N-1}\frac{1}{N}\sum_{l=1}^N\sum_{\array{k=1\\l\neq k}}^N\sum_{\array{j=1 \\j\neq k\\j\neq l}}^N\sum_{\array{i=1\\i\neq j\\i\neq k \\i \neq l}}^Nc(\theta_i)$$

where all the other elements run until $(N-1)$

How can I write a general expression for this series, and can the summation and average of all the elements in the series can be written?

Answer 1

We just need to look at two instances in order to see what's going on.

We obtain \begin{align*} \color{blue}{\sum_{j=1}^N\sum_{{i=1}\atop{i\neq j}}^Nc(\theta_i)} &=\sum_{i=1}^Nc(\theta_i)\sum_{{j=1}\atop{j\neq i}}^N1\\ &\,\,\color{blue}{=(N-1)\sum_{i=1}^Nc(\theta_i)} \end{align*} and \begin{align*} \color{blue}{\sum_{k=1}^N\sum_{{j=1}\atop{j\neq k}}^N\sum_{{i=1}\atop{i\neq j,i\neq k}}^Nc(\theta_i)} &=\sum_{i=1}^Nc(\theta_i)\sum_{{j=1}\atop{j\neq i}}^N\sum_{{i=1}\atop{i\neq j,i\neq k}}^N1\\ &=(N-2)\sum_{i=1}^Nc(\theta_i)\sum_{{j=1}\atop{j\neq i}}^N1\\ &\,\,\color{blue}{=(N-1)(N-2)\sum_{i=1}^Nc(\theta_i)} \end{align*}

We deduce for positive integers $m<N$:

\begin{align*} \frac{(N-m)!}{N!}\sum_{k_0=1}^N \sum_{{k_1=1}\atop{k_1\neq k_j, 0\leq j< 1}}^N\cdots \sum_{{k_m=1}\atop{k_m\neq k_j, 0\leq j<m}}^Nc(\theta_{k_m}) =\frac{N-m}{N}\sum_{k_m=1}^Nc(\theta_{k_m})\tag{1} \end{align*} which can be shown rigorously for instance by using induction by $m\geq 1$.

The general part corresponds to the examples above by \begin{align*} &(j,i)\to(k_0,k_1)\quad &(m=1)\\ &(k,j,i)\to (k_0,k_1,k_2)\quad &(m=2) \end{align*}

We obtain for $m=3$ from (1): \begin{align*} \color{blue}{\frac{(N-3)!}{N!}}&\color{blue}{\sum_{k_0=1}^N \sum_{{k_1=1}\atop{k_1\neq k_j, 0\leq j< 1}}^N \sum_{{k_2=1}\atop{k_2\neq k_j, 0\leq j<2}}^N \sum_{{k_3=1}\atop{k_3\neq k_j, 0\leq j<3}}^N c(\theta_{k_3})}\\ &=\frac{1}{N(N-1)(N-2)}\sum_{k_3=1}^Nc(\theta_{k_3}) \sum_{{k_2=1}\atop{k_2\neq k_{3-j}, 0\leq j< 1}}^N \sum_{{k_1=1}\atop{k_1\neq k_{3-j}, 0\leq j<2}}^N \sum_{{k_0=1}\atop{k_0\neq k_{3-j}, 0\leq j<3}}^N1\\ &=\frac{N-3}{N(N-1)(N-2)}\sum_{k_3=1}^Nc(\theta_{k_3}) \sum_{{k_2=1}\atop{k_2\neq k_{3-j}, 0\leq j< 1}}^N \sum_{{k_1=1}\atop{k_1\neq k_{3-j}, 0\leq j<2}}^N1\\ &=\frac{N-3}{N(N-1)}\sum_{k_3=1}^Nc(\theta_{k_3}) \sum_{{k_2=1}\atop{k_2\neq k_{3-j}, 0\leq j< 1}}^N1\\ &\,\,\color{blue}{=\frac{N-3}{N}\sum_{k_3=1}^Nc(\theta_{k_3})} \end{align*}