Question

Show that $\frac{b}{a+}\frac{b}{a+}\frac{b}{a+}\cdots =b\cdot \frac{{\alpha }^x-{\beta }^x}{{\alpha }^{x+1}-{\beta }^{x+1}}$,
$x$ being the number of components, and $\alpha ,\beta$ the roots of the equation
$k^2-ak-b=0$.

Answer 1

$p=p_1+p_{2}x+p_3{x}^2+....$

$q=q_1+q_{2}x+q_3{x}^2+....$

We see that;-

$p_x=$co efficient of $y^{x-1}$ in $\frac{b}{1-ay-b{y}^2}$

and $q_x=$co efficient of $y^x$ in $\frac{1}{1-ay-b{y}^2}$

If $\alpha ,\beta$ are roots of the equation;-

$k^2-ak-b=0$

$\alpha +\beta =a,\alpha \beta =-b$

$\therefore (bcd+b+d)x^2-(abcd+ab+d-bc+cd)x-(abc+c+a)=0$

If $y=-d+\frac{1}{-c+}\frac{1}{-b+}\frac{1}{-a+}\frac{1}{-d+}...$ by writing $-d,-c,-b,-a$ for a.b.c.d representing in above equation we get;-

$(-abc-c-a)y^2-(abcd+cd+ad-bc+ab)y-(-bcd-b-d)=0$

$(abc+c+a)y^2+(abcd+cd+ad-bc+ab)y-(bcd+b+d)=0$

Now y is the negative root of this equation, $y=-\frac{1}{x}$ we have;-

$(bcd+b+d)z^2-(abcd+ab+ad-bc+cd)z-(abc+c+a)=0$

and $\therefore z=x,y=-\frac{1}{x}$

or $yx=-1$