An equivalent formula for $\sum_{1\le i_1\lt i_2 \dots \lt i_n\le n} a_{i_1} a_{i_2} \dots a_{i_n}$

Question

I know that the following holds:

$\sum_{1\le i\lt j\le n} a_j a_k = \frac12\left(\left(\sum a_i\right)^2-\sum a_i^2\right)$

The question is: does some equivalent formula holds for products of $q>2$ terms, i.e.:

$\sum_{1\le i_1\lt i_2 \dots \lt i_n\le n} a_{i_1} a_{i_2} \dots a_{i_n}$

with $|\{i_1, \dots, i_n \}| = q$

?

Thank you very much for your support!

Answer 1

Define the polynomials $\ e_{k}(a_1,a_2,\ldots,a_n)\ $ by $$e_{k}(a_1,a_2,\ldots,a_n)=\sum_{1\leq i_1<i_2<\cdots i_{k}\leq n}a_{i_1}a_{i_2}\cdots a_{i_k}$$ $$p_{k}(a_1,a_2,\ldots,a_n)=a_{1}^{k}+a_{2}^{k}\cdots+a_{n}^{k}$$ where $1\leq k\leq n$. Then, $$e_{k}=\sum_{j=1}^{k}(-1)^{j-1}\,e_{k-j}(a_1,a_2,\ldots,a_n)\,p_{j}(a_1,a_2,\ldots,a_n)$$ This is what Achille already suggested. Also, you can write $$e_k = (-1)^{k}\sum_{i_1+2i_2+\cdots+ni_n=n}\,\prod_{j=1}^{k}\frac{p_j^{i_j}}{j^{i_j}i_j!}$$