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

Answer: (B)

2

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