I got this formula but can anyone prove it [closed]

Yesterday while doing some questions, I came across a formula that if a wire with uniform resistance R is made into n sided regular polygon then the net resistance between any two corners with $$x-1$$ vertices in between is

$$R_{net} =\frac{Rx(n-x)} {n^2}$$

It's just written in my book with no proof. I have tried proving it but could not. I want the physics behind it rather than just the formula. I would appreciate any kind of help.

closed as off-topic by David Z♦May 10 at 12:35

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• Possible duplicate: physics.stackexchange.com/q/478964/2451 – Qmechanic May 9 at 15:40
• @Satwik Hello Satwik I recommend you to post the proof on this website. – Unique May 9 at 15:44
• Hi Satwik. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic May 9 at 16:55

So you've made an $$n$$ sided polygon out of a wire of resistance $$R$$. Each segment of the polygon therefore has resistance $$R/n$$. If there are $$x - 1$$ vertices between two points, we have basically formed a parallel circuit where one branch has $$x$$ segments and one has $$n - x$$. We then use the formula for a parallel resistance: $$\frac{1}{R_\mathrm{total}} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$R_1 = x R/n$$ and $$R_2 = (n - x) R/n$$. As a result: $$\frac{1}{R_1} + \frac{1}{R_2} = \frac{n}{x R} + \frac{n}{ (n - x) R} = \frac{n(n -x)R + nx R}{(n -x)x R^2} = \frac{n^2}{(n -x )x R}$$
Finally, inverting this yields the following formula: $$R_\mathrm{total} = \frac{ R x (n-x)}{n^2}$$