Prove that every geometric sequence allows for $S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$

Question

$S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$

Is there a way to prove this without expanding everything based on the geometric sum formula? I get lost very easily when trying to solve this conventionally and I feel that I am missing an obvious solution.

Answer 1

Write

$$S_n = a \sum_{k=0}^{n-1} r^k$$

Then

$$S_{2 n} = S_n + r^n S_n$$ $$S_{3 n} = S_{2 n} + r^{2 n}S_n$$

The above result follows from simple algebra.

Answer 2

firsly: $$S_n=a_1\frac{1-r^n}{1-r}$$ so we can say that: $$S_n\left(S_{3n}-S_{2n}\right)=a_1^2\frac{1-r^n}{1-r}\left(\frac{(1-r^{3n})-(1-r^{2n})}{1-r}\right)=a_1^2\frac{(1-r^n)(r^{2n}-r^{3n})}{(1-r)^2}$$ and: $$(S_{2n}-S_n)^2=\left[a_1\frac{1-r^{2n}}{(1-r)}-a_1\frac{1-r^n}{(1-r)}\right]^2=\left[a_1\frac{r^n-r^{2n}}{(1-r)}\right]^2=a_1^2\frac{(r^n-r^{2n})^2}{(1-r)^2}$$ now we just need to show that the top of the fractions are equivalent: $$(r^n-r^{2n})^2=(r^n)^2-2r^nr^{2n}+(r^{2n})^2=r^{2n}-2r^{3n}+r^{4n}$$ $$(1-r^n)(r^{2n}-r^{3n})=r^{2n}-r^{3n}-r^{3n}+r^{4n}=r^{2n}-2r^{3n}+r^{4n}$$ so the two are equivalent