edit: I should have written earlier that:
[MATH]S_2=\frac{8}{9}S_5[/MATH]
because we begin counting with the zeroth term. I apologize for the confusion.
To find \(r\), I would write:
[MATH]\frac{a\left(r^{2+1}-1\right)}{r-1}=\frac{8}{9}\cdot\frac{a\left(r^{5+1}-1\right)}{r-1}[/MATH]
Multiply through by [MATH]\frac{9(r-1)}{a}[/MATH]:
[MATH]9\left(r^{3}-1\right)=8\left(r^{6}-1\right)[/MATH]
[MATH]8r^6-9r^3+1=0[/MATH]
If we recognize this is q quadratic in \(r^3\), we may factor as follows:
[MATH]\left(8r^3-1\right)\left(r^3-1\right)=0[/MATH]
Given that \(|r|<1\), the only real root will come from:
[MATH]8r^3-1=0\implies r=\frac{1}{2}[/MATH]
Now, we should see that:
[MATH]S_{\infty}=\frac{a}{1-r}[/MATH]
Since:
[MATH]\lim_{n\to\infty}S_n=\frac{a}{1-r}[/MATH]
And so what do you get for the infinite sum of the given series?
[/QU
edit: I should have written earlier that:
[MATH]S_2=\frac{8}{9}S_5[/MATH]
because we begin counting with the zeroth term. I apologize for the confusion.
To find \(r\), I would write:
[MATH]\frac{a\left(r^{2+1}-1\right)}{r-1}=\frac{8}{9}\cdot\frac{a\left(r^{5+1}-1\right)}{r-1}[/MATH]
Multiply through by [MATH]\frac{9(r-1)}{a}[/MATH]:
[MATH]9\left(r^{3}-1\right)=8\left(r^{6}-1\right)[/MATH]
[MATH]8r^6-9r^3+1=0[/MATH]
If we recognize this is q quadratic in \(r^3\), we may factor as follows:
[MATH]\left(8r^3-1\right)\left(r^3-1\right)=0[/MATH]
Given that \(|r|<1\), the only real root will come from:
[MATH]8r^3-1=0\implies r=\frac{1}{2}[/MATH]
Now, we should see that:
[MATH]S_{\infty}=\frac{a}{1-r}[/MATH]
Since:
[MATH]\lim_{n\to\infty}S_n=\frac{a}{1-r}[/MATH]
And so what do you get for the infinite sum of the given series?
omg I totally got it now. It turned out I subs in r as a in the denominator in the equation. Thank you so much for your help!!appreciate it!