$\alpha$ & $\beta\;$are +ve roots of $\;x^2-2ax+ab=0\;$ then for $\;n \in N\;$$(0 \lt b \lt a)\;S_{n}=1+2(\large\frac{b}{a})$$+3(\large\frac{b}{a})^{\normalsize 2}$$+\;...+\;n$$(\large\frac{b}{a})^{n-1}\;$ can not exceed .

$(a)\;\frac{\alpha}{\beta}\qquad(b)\;\frac{\beta}{\alpha}\qquad(c)\;|$$\large\frac{\alpha+\beta}{\alpha-\beta}|$$\qquad(d)\;(\large\frac{\alpha+\beta}{\alpha+\beta})^{4}$

Answer : $\;(\large\frac{\alpha+\beta}{\alpha-\beta})^{4}$
Explanation : $\;S_{n} < 1+2(\frac{b}{a})+3(\frac{b}{a})^2+...\infty$
Explanation : $\;\frac{b}{a}\;S_{n} < 1(\frac{b}{a})+2(\frac{b}{a})^2+...\infty$
Subtracting ,
$(1-\frac{b}{a})\;S_{n} < 1+(\frac{b}{a})+(\frac{b}{a})^2+....\;\infty$
$(\large\frac{a-b}{a})S_{n} < \large\frac{1}{(1-\frac{b}{a})}$
$S_{n} < \large\frac{a^2}{(a-b)^2}=(\large\frac{a}{a-b})^2$
$=(\large\frac{4a^2}{4a^2-4ab})^2$
$=(\large\frac{(\alpha+\beta)^2}{(\alpha+\beta)^2-4\alpha\beta})^2$
$=(\large\frac{(\alpha+\beta)^2}{(\alpha-\beta)^2})^2$
$=(\large\frac{\alpha+\beta}{\alpha-\beta})^{4}.$
edited Mar 26, 2014 by yamini.v