I am trying to write a general expression for the equivalent resistance of $n$ parallel resistors. Of course the well-known formula is
$$
R_{eq} = \left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots + \dfrac{1}{R_n}\right)^{-1}.
$$
In the case of two parallel resistors, this can be simplified to a product-over-sum expression:
$$
R_{eq} = \dfrac{R_1R_2}{R_1+R_2}.
$$
When more than 2 resistors are in parallel, it's easier to just use the first equation, but for fun I decided to try to derive a product-over-sum expression for $n$ resistors. This is quite simple to derive but I'm having trouble with how I would express the sum in the denominator. In English, the denominator turns out to be "the sum of all possible combinations of $n – 1$ resistances"; for example, if we have 4 parallel resistors, their equivalent resistance is
$$
R_{eq} = \dfrac{R_1R_2R_3R_4}{R_1R_2R_3 + R_1R_2R_4 + R_1R_3R_4 + R_2R_3R_4}.
$$
The numerator is easy to express using product notation, but like I said I don't know how to express the denominator in the general case of $n$ resistors. I expect it would look something like this:
$$
R_{eq} = \dfrac{\prod\limits_{i=1}^{n}R_i}{\sum\limits_{i=1}^n(\text{something…})}.
$$
Also it would be interesting to generalize the denominator to not just take all possible combinations of $n-1$ quantities, but all possible combinations of any number $k$ of quantities.
Best Answer
Quite trivial if you have seen some conditions/inequalities below the $\sum$ or the $\prod$ symbols
The expression in the denominator you are looking for is $$\sum_{i=1}^n\prod_{\substack{j=1 \\ j\neq i}}^n R_j$$
For the generalization to the 'sum taken $k$ at a time', let us find the sum with respect to the set $S={x_1,x_2,...,x_n}$
We write $$\sum_{\substack{{T\subseteq S} ,\\ {|T|=k}}}\prod_{x\in T} x$$
which would translate to english as 'summing over all k sized subsets of S and multiplying the elements inside them' which is nothing but the sum of the products of any k elements of S