Math Question

Since r is a root of x^3 + 2/5*x - 1 = 0, r^3 + 2/5*r - 1 = 0. Then the sum of the series is

\(\begin{align*} r^2 + \frac{4r - 2/5}{1 - r^3} &= r^2 + \frac{4r - 2/5}{2/5 \cdot r} \\ &= \frac{2/5 \cdot r^3 + 4r - 2/5}{2/5 \cdot r} \\ &= \frac{104/25 \cdot r}{2/5 \cdot r} \\ &= \boxed{\frac{52}{5}}. \end{align*}\)

Let \(r\) be the positive real solution to \(x^3 + \dfrac{2}{5} x - 1 = 0\).
Find the exact numerical value of \(r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb\) .

\(\begin{array}{|rrcll|} \hline & \mathbf{x^3 + \dfrac{2}{5} x - 1} &=& \mathbf{0} \quad | \quad x=r \\\\ & r^3 + \dfrac{2}{5} r - 1 &=& 0 \\ \text{or}& \mathbf{1- r^3} &=& \mathbf{\dfrac{2}{5} r} \\ \hline \end{array} \)

\(\begin{array}{rcrl} s &=& 1*r^2 +& 2*r^5 + 3*r^8 + 4*r^{11} + \dotsb \\ r^3s &=& & 1*r^5 + 2*r^8 + 3*r^{11} + \dotsb \\ \hline s-r^3s &=& 1*r^2 + & 1*r^5+1*r^8+1*r^{11}+ \dotsb \\ \end{array}\)

\(\begin{array}{rcll} s(1-r^3) &=& 1*r^2 + 1*r^5+1*r^8+1*r^{11}+ \dotsb \\ &=& r^2 + r^5+r^8+r^{11}+ \dotsb \\ &=& r^2 \left( 1 + r^3+r^6+r^9+ \dotsb \right) \\ &=& r^2 \left( 1 + \left(r^3\right)^1+\left(r^3\right)^2+\left(r^3\right)^3+ \dotsb \right) \\ && \boxed{ r^3 = q~( \text{common ratio of the infinite geom. sequence })} \\ &=& r^2 \left( 1 + q^1+q^2+q^3+ \dotsb \right) \\ && \boxed{ \text{the sum of an infinite geometric sequence} = \dfrac{1}{1-q} \\ 1 + q^1+q^2+q^3+ \dotsb = \dfrac{1}{1-q} } \\ &=& r^2 \dfrac{1}{1-q} \quad | \quad q=r^3 \\\\ &=& r^2 \dfrac{1}{1-r^3} \\\\ s(1-r^3) &=& \dfrac{r^2}{1-r^3} \\\\ s &=& \dfrac{r^2}{\left(1-r^3\right)^2}\quad | \quad 1-r^3 = \dfrac{2}{5}r \\\\ s &=& \dfrac{r^2}{\left(\dfrac{2}{5}r\right)^2} \\\\ s &=& \dfrac{r^2}{\left(\dfrac{2}{5}\right)^2r^2} \\\\ s &=& \dfrac{1}{\left(\dfrac{2}{5}\right)^2} \\\\ s &=& \left(\dfrac{5}{2}\right)^2 \\\\ \mathbf{s} &=& \mathbf{ \dfrac{25}{4} } \\ \end{array}\)

\(\mathbf{ r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb = \dfrac{25}{4} }\)