# Find the geometric power series that represents the given function. \frac{1}{9 + x} centered at x= 2

## Question:

Find the geometric power series that represents the given function.

{eq}\frac{1}{9 + x} {/eq} centered at x= 2

## Calculation of Geometric Power Series:

If {eq}k {/eq} is any number and {eq}\left| x \right| < 1 {/eq}, then the binomial series expansion is given as:

{eq}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} k \\ n \\ \end{array}} \right){x^n}} = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..... {/eq}

Using the concept of binomial series expansion, we can expand any function to the form of a geometric power series.

Here, in this case, the given function is expanded as:

{eq}\eqalign{ f\left( x \right) = \frac{1}{{9 + x}} & \Rightarrow f\left( x \right) = \frac{1}{{11 + x - 2}} \cr & \Rightarrow f\left( x \right) = \frac{1}{{11}}\left( {\frac{1}{{1 + \left( {\frac{{x - 2}}{{11}}} \right)}}} \right) \cr & \Rightarrow f\left( x \right) = \frac{1}{{11}}{\left( {1 + \left( {\frac{{x - 2}}{{11}}} \right)} \right)^{ - 1}} \cr & \Rightarrow f\left( x \right) = \frac{1}{{11}}\left( {1 - \left( {\frac{{x - 2}}{{11}}} \right) + {{\left( {\frac{{x - 2}}{{11}}} \right)}^2} - {{\left( {\frac{{x - 2}}{{11}}} \right)}^3} + {{\left( {\frac{{x - 2}}{{11}}} \right)}^4} - ....} \right)\,\,\,\left[ {{\text{Using the power series }}{{\left( {1 + x} \right)}^{ - 1}} = 1 - x + {x^2} - {x^3} + {x^4} - .....} \right] \cr & \Rightarrow f\left( x \right) = \frac{1}{{11}}\sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{{x - 2}}{{11}}} \right)}^n}} \,\,\,\,\left[ {{\text{Using the concept of power series}}} \right] \cr & \Rightarrow f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{11}^{n + 1}}}}{{\left( {x - 2} \right)}^n}} \cr} {/eq}

This is the required geometric power series of the given function. 