Answer:
[tex]\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........[/tex]
The interval of convergence is:[tex](-\frac{2}{3},\frac{16}{3})[/tex]
Step-by-step explanation:
Given
[tex]f(x)= \frac{9}{3x+ 2}[/tex]
[tex]c = 6[/tex]
The geometric series centered at c is of the form:
[tex]\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.[/tex]
Where:
[tex]a \to[/tex] first term
[tex]r - c \to[/tex] common ratio
We have to write
[tex]f(x)= \frac{9}{3x+ 2}[/tex]
In the following form:
[tex]\frac{a}{1 - r}[/tex]
So, we have:
[tex]f(x)= \frac{9}{3x+ 2}[/tex]
Rewrite as:
[tex]f(x) = \frac{9}{3x - 18 + 18 +2}[/tex]
[tex]f(x) = \frac{9}{3x - 18 + 20}[/tex]
Factorize
[tex]f(x) = \frac{1}{\frac{1}{9}(3x + 2)}[/tex]
Open bracket
[tex]f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}[/tex]
Rewrite as:
[tex]f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}[/tex]
Collect like terms
[tex]f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}[/tex]
Take LCM
[tex]f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}[/tex]
[tex]f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}[/tex]
So, we have:
[tex]f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}[/tex]
By comparison with: [tex]\frac{a}{1 - r}[/tex]
[tex]a = 1[/tex]
[tex]r = -\frac{1}{3}x + \frac{7}{9}[/tex]
[tex]r = -\frac{1}{3}(x - \frac{7}{3})[/tex]
At c = 6, we have:
[tex]r = -\frac{1}{3}(x - \frac{7}{3}+6-6)[/tex]
Take LCM
[tex]r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)[/tex]
r = -\frac{1}{3}(x + \frac{11}{3}+6-6)
So, the power series becomes:
[tex]\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n[/tex]
Substitute 1 for a
[tex]\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n[/tex]
[tex]\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n[/tex]
Substitute the expression for r
[tex]\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n[/tex]
Expand
[tex]\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n][/tex]
Further expand:
[tex]\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................[/tex]
The power series converges when:
[tex]\frac{1}{3}|x - \frac{7}{3}| < 1[/tex]
Multiply both sides by 3
[tex]|x - \frac{7}{3}| <3[/tex]
Expand the absolute inequality
[tex]-3 < x - \frac{7}{3} <3[/tex]
Solve for x
[tex]\frac{7}{3} -3 < x <3+\frac{7}{3}[/tex]
Take LCM
[tex]\frac{7-9}{3} < x <\frac{9+7}{3}[/tex]
[tex]-\frac{2}{3} < x <\frac{16}{3}[/tex]
The interval of convergence is:[tex](-\frac{2}{3},\frac{16}{3})[/tex]
