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# A Visual Proof for Geometric Series

I personally love the idea of Proofs Without Words! I like how simple, elegant and creative they can be. So here is another one for all you proof enthusiasts. Link

Expansion

In this proof, we see that $$\frac{1}{1-r}$$ = $$1 + r + r^2 + r^3 ...$$.

$$\triangle TPS$$ is cut into multiple similar triangles with a ratio r. $${ST}$$ is $$1 + r + r^2 + r^3 ...$$. For the purposes of this proof, let the point where line $${QR}$$ meets $$ST$$ be $$A$$.

$$\angle PRQ = \angle TRA$$ because they're vertical angles. $$\angle PQR = \angle TAR$$ since they are both right angles. Thus $$\triangle PRQ \sim \triangle TRA$$ by $$AA$$ Similarity. Since $$\triangle TPS \sim \triangle TRA$$, $$\triangle PRQ \sim \triangle TPS$$.

Tinkering with our ratios, we get $$\frac{PQ}{QR} = \frac{TS}{SP}$$

$$\frac{1}{1-r} = \frac{1 + r + r^2 + r^3 ...}{1}$$

Q.E.D

Note by Sherry Sarkar
2 years, 9 months ago

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This is awesome! Although perhaps it is worth noting that $$|r| < 1$$ for this to work. I wonder if I could think up a similar geometric proof for the more general formula: $$1 + r + r^{2} +... = \dfrac{1-r^{n}}{1-r}$$... Hmm... · 2 years, 9 months ago

Thank you for the feedback, I forgot to add that detail! · 2 years, 9 months ago

Elegant Maths... that's what i call it · 2 years, 9 months ago

It's really amazing, · 2 years, 9 months ago

Thanks! It's my first post, so I'm on the lines here. :D · 2 years, 9 months ago

Nice one for the first post. Keep posting! · 2 years, 9 months ago

We will get the same answer if we extend the dotted triangle towards the right:

If we extend upto r^2, we get $(1+r)/(1- r^{2} )$ which can be simplified to get the same answer.

If we extend upto r^3, we get $(1+r+r^{2})/(1-r^{3})$ which again can be simplified to get the same result. And so on...... · 2 years, 7 months ago

Wow. Nice proof! · 2 years, 9 months ago