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 $3 \,x^{2}+6\, x$ 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, $\left(1-\frac{2}{5}\right)^{-1 / 2}=1+\frac{1}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{1}{5}\right)^{2}+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{1}{5}\right)^{3}+\ldots$
$\Rightarrow 1+x=\left(1-\frac{2}{5}\right)^{-1 / 2} \Rightarrow 1+x=\left(\frac{3}{5}\right)^{-1 / 2}$
$\Rightarrow 1+x=\left(\frac{5}{3}\right)^{1 / 2}$
On squaring both sides, we get
$ (1+x)^{2} =\frac{5}{3}$
$\Rightarrow 1+2\, x+x^{2} =\frac{5}{3} $
$\Rightarrow 3+6 \,x+3 x^{2} =5 $
$\Rightarrow 3 \,x^{2}+6 \,x=2$