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Markus Scheuer
Markus Scheuer
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It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)}\tag{2} \end{align*} from where you can continue in the way you have described.

[2024-11-18] With respect to OPs comment some more details. We can continue from (2) with \begin{align*} \frac{1}{\sqrt{21}}&\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\cdot\frac{1}{1-\frac{x}{x_1}}+\frac{1}{x_2}\cdot\frac{1}{1-\frac{x}{x_2}}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\sum_{n=0}^{\infty}\left(\frac{x}{x_1}\right)^n +\frac{1}{x_2}\sum_{n=0}^{\infty}\left(\frac{x}{x_2}\right)^n\right)\\ &=\frac{1}{\sqrt{21}}\sum_{n=0}^{\infty}\left(\frac{1}{{x_2}^{n+1}} -\frac{1}{{x_1}^{n+1}}\right)x^n\tag{3} \end{align*} We know from OP that \begin{align*} x_1=-\frac{1}{6}\left(3+\sqrt{21}\right)\qquad\qquad x_2=-\frac{1}{6}\left(3-\sqrt{21}\right) \end{align*} from which we derive by also rationalizing the denominator \begin{align*} \frac{1}{x_1}&=-\frac{6}{3+\sqrt{21}}=\frac{3-\sqrt{21}}{2}\\ \frac{1}{x_2}&=-\frac{6}{3-\sqrt{21}}=\frac{3+\sqrt{21}}{2}\\ \end{align*} Putting this into (3) we finally obtain \begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x++ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*}\begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x+ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*} in accordance with the claim. The last line was calculated with some help of Wolfram Alpha.

It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)}\tag{2} \end{align*} from where you can continue in the way you have described.

[2024-11-18] With respect to OPs comment some more details. We can continue from (2) with \begin{align*} \frac{1}{\sqrt{21}}&\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\cdot\frac{1}{1-\frac{x}{x_1}}+\frac{1}{x_2}\cdot\frac{1}{1-\frac{x}{x_2}}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\sum_{n=0}^{\infty}\left(\frac{x}{x_1}\right)^n +\frac{1}{x_2}\sum_{n=0}^{\infty}\left(\frac{x}{x_2}\right)^n\right)\\ &=\frac{1}{\sqrt{21}}\sum_{n=0}^{\infty}\left(\frac{1}{{x_2}^{n+1}} -\frac{1}{{x_1}^{n+1}}\right)x^n\tag{3} \end{align*} We know from OP that \begin{align*} x_1=-\frac{1}{6}\left(3+\sqrt{21}\right)\qquad\qquad x_2=-\frac{1}{6}\left(3-\sqrt{21}\right) \end{align*} from which we derive by also rationalizing the denominator \begin{align*} \frac{1}{x_1}&=-\frac{6}{3+\sqrt{21}}=\frac{3-\sqrt{21}}{2}\\ \frac{1}{x_2}&=-\frac{6}{3-\sqrt{21}}=\frac{3+\sqrt{21}}{2}\\ \end{align*} Putting this into (3) we finally obtain \begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x++ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*} in accordance with the claim. The last line was calculated with some help of Wolfram Alpha.

It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)}\tag{2} \end{align*} from where you can continue in the way you have described.

[2024-11-18] With respect to OPs comment some more details. We can continue from (2) with \begin{align*} \frac{1}{\sqrt{21}}&\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\cdot\frac{1}{1-\frac{x}{x_1}}+\frac{1}{x_2}\cdot\frac{1}{1-\frac{x}{x_2}}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\sum_{n=0}^{\infty}\left(\frac{x}{x_1}\right)^n +\frac{1}{x_2}\sum_{n=0}^{\infty}\left(\frac{x}{x_2}\right)^n\right)\\ &=\frac{1}{\sqrt{21}}\sum_{n=0}^{\infty}\left(\frac{1}{{x_2}^{n+1}} -\frac{1}{{x_1}^{n+1}}\right)x^n\tag{3} \end{align*} We know from OP that \begin{align*} x_1=-\frac{1}{6}\left(3+\sqrt{21}\right)\qquad\qquad x_2=-\frac{1}{6}\left(3-\sqrt{21}\right) \end{align*} from which we derive by also rationalizing the denominator \begin{align*} \frac{1}{x_1}&=-\frac{6}{3+\sqrt{21}}=\frac{3-\sqrt{21}}{2}\\ \frac{1}{x_2}&=-\frac{6}{3-\sqrt{21}}=\frac{3+\sqrt{21}}{2}\\ \end{align*} Putting this into (3) we finally obtain \begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x+ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*} in accordance with the claim. The last line was calculated with some help of Wolfram Alpha.

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Markus Scheuer
Markus Scheuer
  • 111.1k
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It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)} \end{align*}\begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)}\tag{2} \end{align*} from where you can continue in the way you have described.

[2024-11-18] With respect to OPs comment some more details. We can continue from (2) with \begin{align*} \frac{1}{\sqrt{21}}&\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\cdot\frac{1}{1-\frac{x}{x_1}}+\frac{1}{x_2}\cdot\frac{1}{1-\frac{x}{x_2}}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\sum_{n=0}^{\infty}\left(\frac{x}{x_1}\right)^n +\frac{1}{x_2}\sum_{n=0}^{\infty}\left(\frac{x}{x_2}\right)^n\right)\\ &=\frac{1}{\sqrt{21}}\sum_{n=0}^{\infty}\left(\frac{1}{{x_2}^{n+1}} -\frac{1}{{x_1}^{n+1}}\right)x^n\tag{3} \end{align*} We know from OP that \begin{align*} x_1=-\frac{1}{6}\left(3+\sqrt{21}\right)\qquad\qquad x_2=-\frac{1}{6}\left(3-\sqrt{21}\right) \end{align*} from which we derive by also rationalizing the denominator \begin{align*} \frac{1}{x_1}&=-\frac{6}{3+\sqrt{21}}=\frac{3-\sqrt{21}}{2}\\ \frac{1}{x_2}&=-\frac{6}{3-\sqrt{21}}=\frac{3+\sqrt{21}}{2}\\ \end{align*} Putting this into (3) we finally obtain \begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x++ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*} in accordance with the claim. The last line was calculated with some help of Wolfram Alpha.

It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)} \end{align*} from where you can continue in the way you have described.

It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)}\tag{2} \end{align*} from where you can continue in the way you have described.

[2024-11-18] With respect to OPs comment some more details. We can continue from (2) with \begin{align*} \frac{1}{\sqrt{21}}&\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\cdot\frac{1}{1-\frac{x}{x_1}}+\frac{1}{x_2}\cdot\frac{1}{1-\frac{x}{x_2}}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\sum_{n=0}^{\infty}\left(\frac{x}{x_1}\right)^n +\frac{1}{x_2}\sum_{n=0}^{\infty}\left(\frac{x}{x_2}\right)^n\right)\\ &=\frac{1}{\sqrt{21}}\sum_{n=0}^{\infty}\left(\frac{1}{{x_2}^{n+1}} -\frac{1}{{x_1}^{n+1}}\right)x^n\tag{3} \end{align*} We know from OP that \begin{align*} x_1=-\frac{1}{6}\left(3+\sqrt{21}\right)\qquad\qquad x_2=-\frac{1}{6}\left(3-\sqrt{21}\right) \end{align*} from which we derive by also rationalizing the denominator \begin{align*} \frac{1}{x_1}&=-\frac{6}{3+\sqrt{21}}=\frac{3-\sqrt{21}}{2}\\ \frac{1}{x_2}&=-\frac{6}{3-\sqrt{21}}=\frac{3+\sqrt{21}}{2}\\ \end{align*} Putting this into (3) we finally obtain \begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x++ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*} in accordance with the claim. The last line was calculated with some help of Wolfram Alpha.

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Markus Scheuer
Markus Scheuer
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It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)} \end{align*} from where you can continue in the way you have described.