Question

If $x=\frac{1}{5}+\frac{1.3}{5.10}+\frac{\mathrm{1.3.5}}{\mathrm{5.10.15}}+....\infty$ then $3x^2+6x$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$1+x=1+\frac{1}{5}+\frac{1.3}{2!}{\left(\begin{array}{c}\frac{1}{5}\end{array}\right)}^2+\frac{\mathrm{1.3.5}}{3!}{\left(\begin{array}{c}\frac{1}{5}\end{array}\right)}^3$
$(1+x{)}^{-}\frac{p}{q}=1+\frac{p(p+q)}{2!}{\left(\begin{array}{c}\frac{x}{q}\end{array}\right)}^2+.....$
$\frac{x}{q}=\frac{1}{5}$
$\Rightarrow p=1,p+q=3,\Rightarrow q=2$
$\therefore x=\frac{q}{5}=\frac{2}{5}$
We get $1+x={\left(\begin{array}{c}\frac{5}{3}\end{array}\right)}^{\frac{1}{2}}$
$x^2+2x+1=\frac{5}{3}$
$3x^2+6x=2$