1 / R total = 1 / R1 + 1 / r2 + +How to simplify 1 / RN? This is the law of resistance in parallel circuit. I remember a simpler formula. How to simplify it?

1 / R total = 1 / R1 + 1 / r2 + +How to simplify 1 / RN? This is the law of resistance in parallel circuit. I remember a simpler formula. How to simplify it?

It can't be simplified
Generally two resistors have a simple form
R1*R2/（R1+R2）

The formula of resistance parallel connection is: 1 / r = 1 / R1 + 1 / r2 +... + 1 / RN
For two resistors, there are two terms on the right side of the equation. Divide them, add them, and count them down again. Thus, r = r1r2 / (R1 + R2)
If R1 is 6 ohm and R2 is 3 ohm, the result calculated by 1 / r = 1 / R1 + 1 / r2 +... + 1 / RN is 1 / 2 ohm, and the result calculated by r = r1r2 / (R1 + R2) is 2 ohm. How can I explain this

If R1 is 6 ohm, R2 is 3 ohm, 1 / r = 1 / R1 + 1 / r2 +... + 1 / RN, the result is 1 / 2
That is: 1 / r = 1 / 2
So: r = 2 ohm
It is consistent with the formula r = R 1R 2 / (R 1 + R 2) that the result is 2 ohm

It is proved that in a series circuit, the total resistance is equal to the sum of the resistances, that is, R total = R1 + R2 + +RN (hint: in fact, we only need to prove that r = R1 + R2, but we need to further explain that multiple resistors in series can be regarded as the result of the first several resistors in series and then the last one in series)

It is proved that the current in series circuit is equal everywhere (axiom), r = u / I
The power supply voltage is equal to the sum of the voltages of each section (axiom),
R total = (U1 + U2 +...) +Un）/I=U1/I+U2/I+…… +Un/I=R1+R2+…… +Rn