Answer:
[tex]f(x) = \sum\limits^{\infty}_{n=0} \frac{-(1)^n\cdot x^n}{5^{n+1}}[/tex]
Step-by-step explanation:
Given
[tex]f(x) = \frac{1}{5 + x}[/tex]
Required
The power series centered at [tex]x = 0[/tex]
We have:
[tex]f(x) = \frac{1}{5 + x}[/tex]
Factor out 5 from the denominator
[tex]f(x) = \frac{1}{5(1 + \frac{x}{5})}[/tex]
Rewrite as:
[tex]f(x) = \frac{\frac{1}{5}}{(1 + \frac{x}{5})}[/tex]
Further, rewrite as:
[tex]f(x) = \frac{1}{5}(1 + \frac{x}{5})^{-1}[/tex]
Expand the bracket
[tex]f(x) = \frac{1}{5}(1 - \frac{x}{5} + (\frac{x}{5})^2 - (\frac{x}{5})^3+..........)[/tex]
Evaluate all exponents
[tex]f(x) = \frac{1}{5}(1 - \frac{x}{5} + \frac{x^2}{25} - \frac{x^3}{125}+......)[/tex]
Open brackets
[tex]f(x) = \frac{1}{5} - \frac{x}{5^2} + \frac{x^2}{5^3} - \frac{x^3}{5^4}+......[/tex]
Notice the pattern as:
[tex]f(x) = \frac{1}{5} - \frac{x}{5^2} + \frac{x^2}{5^3} - \frac{x^3}{5^4}+......\± \frac{x^n}{5^{n+1}}[/tex]
So, the power series is:
[tex]f(x) = \sum\limits^{\infty}_{n=0} \frac{-(1)^n\cdot x^n}{5^{n+1}}[/tex]
