MHB Where did I go wrong in simplifying this algebraic expression?

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The discussion revolves around the simplification of the algebraic expression $\alpha_nr^n + \beta_nr^{-n}$, where initial formulations for $\alpha_n$ and $\beta_n$ were incorrectly derived. The user initially calculated $\alpha_n$ and $\beta_n$ but ended up with an expression that did not match the expected solution. By rearranging $\alpha_n$ and $\beta_n$ to have a common denominator, the user successfully simplified the expression to match the correct form. The final result confirms that the error lay in the initial handling of the denominators. The thread concludes with the user marking the issue as resolved.
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$\alpha_nr^n + \beta_nr^{-n}$

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.
 
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dwsmith said:
$\alpha_nr^n + \beta_nr^{-n}$

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}
 
Sudharaka said:
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.
 
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