<s>[INST] Confused about sum of G.P. formula The sum of $n$ terms of the G.P. $a, ar, ar^2, ar^3, \ldots, ar^{n-1}$ is given by $a\dfrac{r^n-1}{r-1}$.
Now consider these two progressions:
1) $r^2, r^4, r^6, ..., r^{2n}$
2) $r, r^3, r^5, ..., r^{2n-1}$
Both of these have $n$ terms. Therefore, $S_{n1} = r^2 \dfrac{(r^2)^n-1}{r-1} = r^2 \dfrac{r^{2n}-1}{r-1}$. Also, $S_{n2} = r \dfrac{r^{2n}-1}{r-1}$. It seems that no matter what $r$ be, the sum of even powers will be $r$ times the sum of odd powers. However, this is not true as a simple example shows:
Consider $r = 0.5$. Now, $r^2 + r^4 + r^6 = 0.328125$, while $r + r^3 = 0.625$, and $ 0.328125 \neq 0.625 \times 0.5$
Where is the error?
 [/INST] The last term of the second series is $r^{2n-1}=r\cdot r^{2n-2}=r\cdot (r^2)^{n-1}$, and the ratio of consecutive terms is $r^2$, not $r$, so the sum formula should be
$$S_{2n}=r\cdot\frac{r^{2n}-1}{r^2-1}\;.$$
The ratio in the first series is also $r^2$, so the sum formula yields
$$S_{1n}=r^2\cdot\frac{r^{2n}-1}{r^2-1}\;.$$
This says that $S_{1n}=rS_{2n}$, which is clearly correct, since
$$r^2+r^4+\ldots+r^{2n}=r\left(r+r^3+\ldots+r^{2n-1}\right)\;.$$
Your example doesn’t match your formulas: for $n=3$ you should be comparing $r^2+r^4+r^6$ with $r+r^3+r^5$, not with $r+r^3$.
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