Answer:
[tex]\mathbf{g(x) = \sum \limits^{\infty}_{n=0} (-1 + (\dfrac{-1}{6})^n)x^n }[/tex]
and the interval of the convergence is (-1, 1)
Step-by-step explanation:
To find a power series for the function, centered at c.
[tex]g(x) = \dfrac{7x}{x^2 +5x-6} ,\ c = 0[/tex]
If we factorize the denominator, we have:
[tex]g(x) = \dfrac{7x}{(x +6)(x-1)}[/tex]
[tex]g(x)= \dfrac{1}{x-1}+\dfrac{6}{x+6}[/tex]
Thus;
[tex]g(x)= \dfrac{-1}{1-x}+\dfrac{1}{1+\dfrac{x}{6}}[/tex]
[tex]g(x)= \dfrac{-1}{1-x}+\dfrac{1}{1-(-\dfrac{x}{6})}[/tex]
[tex]g(x) = - \sum \limits^{\infty}_{n=0} x^n + \sum \limits^{\infty}_{n=0} x^n(\dfrac{-x}{6})^n \ \ if \ \ |x| < 1 \ \ and \ \ |\dfrac{x}{6}< 1[/tex]
[tex]g(x) = \sum \limits^{\infty}_{n=0} (-1 + (\dfrac{-1}{6})^n)x^n , \ if |x|<1[/tex]
[tex]\mathbf{g(x) = \sum \limits^{\infty}_{n=0} (-1 + (\dfrac{-1}{6})^n)x^n }[/tex]
and the interval of the convergence is (-1, 1)
