Question

If n be a root of the equation
$x^2(1-ab)-x(a^2+b^2)-(1+ab)=0.$
Prove that $H_1-H_n=ab(a-b)$

Answer 1

$H_n-H_1=\frac{(n+1)ab}{na+b}-\frac{(n+1)ab}{(nb+a)}$
$=\frac{(n+1)ab[nb+a-na-b]}{n^{2}ab+n(a^2+b^2)+ab}$
$=\frac{(n+1)ab(1-n)(a-b)}{n^2-1}$
= -ab (a - b), by (1) below
$\therefore H_1-H_n=ab(a-b).$
Since n is a root of the given equation
$\therefore n^2(1-ab)-n(a^2+b^2)-(1+ab)=0$
or $n^2-1=n^{2}ab+n(a^2+b^2)+ab$