Question

If $x=\frac{1}{5}+\frac{1\cdot 3}{5\cdot 10}+\frac{1\cdot 3\cdot 5}{5\cdot 10\cdot 5}+\cdots \infty$,
then $3x^2+6x$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

We have,
$x=\frac{1}{5}+\frac{1\cdot 3}{5\cdot 10}+\frac{1\cdot 3\cdot 5}{5\cdot 10\cdot 5}+\cdots$
$=\frac{1}{5}+\frac{1\cdot 3}{2\times 1}{\left(\frac{1}{5}\right)}^2+\frac{1\cdot 3\cdot 5}{3\times 2\times 1}{\left(\frac{1}{5}\right)}^3+\cdots$
$=\frac{1}{5}+\frac{1\cdot 3}{2!}{\left(\frac{1}{5}\right)}^2+\frac{1\cdot 3\cdot 5}{3!}{\left(\frac{1}{5}\right)}^3+\cdots$
On adding 1 both sides, we get
$1+x=1+\frac{1}{5}+\frac{1\cdot 3}{2!}{\left(\frac{1}{5}\right)}^2+\frac{1\cdot 3\cdot 5}{3!}{\left(\frac{1}{5}\right)}^3+\cdots$
Now, ${\left(1-\frac{2}{5}\right)}^{-1/2}=1+\frac{1}{5}+\frac{1\cdot 3}{2!}{\left(\frac{1}{5}\right)}^2+\frac{1\cdot 3\cdot 5}{3!}{\left(\frac{1}{5}\right)}^3+\dots$
$\Rightarrow 1+x={\left(1-\frac{2}{5}\right)}^{-1/2}\Rightarrow 1+x={\left(\frac{3}{5}\right)}^{-1/2}$
$\Rightarrow 1+x={\left(\frac{5}{3}\right)}^{1/2}$
On squaring both sides, we get
$(1+x{)}^2=\frac{5}{3}$
$\Rightarrow 1+2x+x^2=\frac{5}{3}$
$\Rightarrow 3+6x+3x^2=5$
$\Rightarrow 3x^2+6x=2$