Question

If $z=\frac{1}{3}+\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\dots$ then

• A. $z^2-2z+2=0$
• B. $z^2-2z-2=0$
• C. $z^2+2z-2=0$
• D. $z^2+2z+2=0$

Answer 1

Answer: (C)

$z^2+2z-2=0$

Adding 1 on both the sides we get
$z+1=1+\frac{1}{3}...\infty$
$(1-x{)}^{-n}=1-n(-x)+\frac{-n(-n-1)x^2}{2!}...\infty$
$=1+nx+\frac{n(n+1)x^2}{2!}...\infty$
Comparing the coefficients with that given in the question we get.
$nx=\frac{1}{3}$ ...(i)
$\frac{n(n+1)x^2}{2}=\frac{1}{6}$ ...(ii)
Substituting the value of nx from Eq i in Eq ii we get
$n{x}^2(n+1)=\frac{1}{3}$
$nx(n+1)x=\frac{1}{3}$
$(n+1)x=1$
$nx+x=1$
$\frac{1}{3}=1-x$
$x=\frac{2}{3}$
Since $nx=\frac{1}{3}$
$n=\frac{1}{2}$
Substituting in the original equation the values of n and x we get
$z+1=(1-\frac{2}{3}{)}^{\frac{-1}{2}}$
$z+1=(\frac{1}{3}{)}^{\frac{-1}{2}}$
$z+1=3^{\frac{1}{2}}$
$(z+1{)}^2=3$
$z^2+2z+1=3$
$z^2+2z-2=0$