# [SOLVED]Algebraic Simplification

#### dwsmith

##### Well-known member
$\alpha_nr^n + \beta_nr^{-n}$

We know that
\begin{alignat*}{3}
\alpha_na^n+\beta_na^{-n} & = & A_n\\
\alpha_nb^n+\beta_nb^{-n} & = & 0
\end{alignat*}
So I ended up with
$$\alpha_n = \frac{-A_n}{b^{2n}/a^n-a^n}$$
and
$$\beta_n = \frac{A_n}{1/a^n-a^n/b^{2n}}$$

When I plug them in, I obtain
$$\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}/a^n-r^{2n}}{b^{2n}-a^{2n}}$$

However, the solution is supposed to be
$$\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}$$

I cannot find my error.

#### Sudharaka

##### Well-known member
MHB Math Helper
$\alpha_nr^n + \beta_nr^{-n}$

We know that
\begin{alignat*}{3}
\alpha_na^n+\beta_na^{-n} & = & A_n\\
\alpha_nb^n+\beta_nb^{-n} & = & 0
\end{alignat*}
So I ended up with
$$\alpha_n = \frac{-A_n}{b^{2n}/a^n-a^n}$$
and
$$\beta_n = \frac{A_n}{1/a^n-a^n/b^{2n}}$$

When I plug them in, I obtain
$$\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}/a^n-r^{2n}}{b^{2n}-a^{2n}}$$

However, the solution is supposed to be
$$\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}$$

I cannot find my error.
I think by rearranging the answers for $$\alpha_n$$ and $$\beta_n$$ to have a common denominator may help in the algebraic simplification.

$\alpha_n=\frac{-A_{n}a^n}{b^{2n}-a^{2n}}$

$\beta_n=\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}$

\begin{eqnarray}

\therefore \alpha_nr^n + \beta_nr^{-n}&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\frac{A_{n}a^nb^{2n}-A_{n}a^nr^{2n}}{r^n(b^{2n}-a^{2n})}\\

&=&\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}

\end{eqnarray}

#### dwsmith

##### Well-known member
I think by rearranging the answers for $$\alpha_n$$ and $$\beta_n$$ to have a common denominator may help in the algebraic simplification.

$\alpha_n=\frac{-A_{n}a^n}{b^{2n}-a^{2n}}$

$\beta_n=\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}$

\begin{eqnarray}

\therefore \alpha_nr^n + \beta_nr^{-n}&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\frac{A_{n}a^nb^{2n}-A_{n}a^nr^{2n}}{r^n(b^{2n}-a^{2n})}\\

&=&\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}

\end{eqnarray}
I marked the thread solved a little bit ago.