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

$1+x=1+\frac{1}{5}+\frac{1.3}{2!}(\frac{1}{5}{)}^2+\frac{\mathrm{1.3.5}}{3!}(\frac{1}{5}{)}^3$
$(1+x{)}^{-p}{/}_q=1+\frac{p(p+q)}{2!}(\frac{x}{q}{)}^2+.....$
$\frac{x}{q}=\frac{1}{5}$
$\Rightarrow p=1,p+q=2,\Rightarrow q=2$
$\therefore x=\frac{q}{5}=\frac{2}{5}$
We get $1+x=(\frac{5}{3}{)}^1{/}_2$
$x^2+2x+1=\frac{5}{3}$
$3x^2+6x=2$