Question

If $x=\frac{1}{3}+\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+....\infty$ , then $3x^2+6x=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$1+x=1+\frac{1}{3}+\frac{1.3}{2!}{\left(\frac{1}{3}\right)}^2+\frac{\mathrm{1.3.5}}{3!}{\left(\frac{1}{3}\right)}^3$
${(1+x)}^{-\frac{p}{q}}=1+\frac{p(p+q)}{2!}{\left(\frac{x}{q}\right)}^2+....$
$\frac{x}{q}=\frac{1}{3}$
$\Rightarrow p=1,p+q=3,\Rightarrow q=2$
$\therefore x=\frac{q}{3}=\frac{2}{3}$
We get $1+x={\left(\frac{5}{3}\right)}^{\frac{1}{2}}$
$x^2+2x+1=\frac{5}{3}$

$3x^2+6x=2$