Question

Let ' $X$ ' denotes the value of the product
$\left(1+a+a^2+a^3+\dots \dots \infty \right)\left(1+b+b^2+b^3+\dots \dots \infty \right)$
where ${}^{\prime }a$ ' and ' $b$ ' are the roots of the quadratic equation $11x^2-4x-2=0$ and ' $Y$ ' denotes the numerical value of the infinite series
${\left({\log }_{c}2\right)}^0\left({\log }_c{5}^{4^0}\right)+{\left({\log }_{c}2\right)}^1\left({\log }_c{5}^{4^1}\right)+{\left({\log }_{c}2\right)}^2\left({\log }_c{5}^{4^2}\right)+{\left({\log }_{c}2\right)}^3\left({\log }_c{5}^{4^3}\right)+\dots \dots \infty$
where $c=2000$. If the value of $(XY)$ can be expressed as rational $\frac{p}{q}$ (where $p,q\in N)$, find $(p+q)$.

Answer 1

Answer: 26

$X=\frac{1}{1-a}\cdot \frac{1}{1-b}=\frac{1}{1-(a+b)+ab}$ where $a+b=\frac{4}{11}$ and $ab=-\frac{2}{11}$
$\therefore X={\left(1-\frac{4}{11}-\frac{2}{11}\right)}^{-1}={\left(1-\frac{6}{11}\right)}^{-1}\Rightarrow X=\frac{11}{5}$

Roots are
$\alpha ,\beta =\frac{4\pm \sqrt{16+88}}{22}$
$=\frac{4\pm \sqrt{104}}{22}=\frac{2\pm \sqrt{26}}{11}$

$\text{ for }Y:a={\log }_{b}5;r=\frac{4{\log }_{b}5\cdot {\log }_{b}2}{{\log }_{b}5}=4{\log }_{b}2$
$\therefore Y=\frac{{\log }_{b}5}{1-4{\log }_{b}2}=\frac{{\log }_{b}5}{{\log }_{b}b-{\log }_{b}16}=\frac{{\log }_{b}5}{{\log }_b\frac{b}{16}}=\frac{{\log }_{2000}5}{{\log }_{2000}125}={\log }_{125}5=\frac{1}{3}\Rightarrow Y=\frac{1}{3}$
$\Rightarrow XY=\frac{11}{5}\cdot \frac{1}{3}=\frac{11}{15}$