Question

If the sum to infinity of the series, $1+4x+7x^2+10x^3+........,$ is $35/16$, where $|x|<1$, then $x$ equals to

• A. $19/7$
• B. $1/5$
• C. $1/4$
• D. None of these

Answer 1

Answer: (B)

$S=1+4x+7x^2+10x^3+........$
$x.S=x+4x^2+7x^3+........$
Subtract
$S(1-x)=1+3x+3x^2+3x^3+.......$
$S\left(1-x\right)=1+3x\left(\frac{1}{1-x}\right),\because \left|x\right|<1$
$S=\frac{1+2x}{{\left(1-x\right)}^2}$
Given : $\frac{1+2x}{{\left(1-x\right)}^2}=\frac{35}{16}$
$\Rightarrow 16+32x=35+35x^2-70x$
$\Rightarrow 35x^2-102x+19=0$
$\Rightarrow 35x^2-7x-95x+19=0$
$\Rightarrow 7x\left(5x-1\right)-19\left(5x-1\right)=0$
$\Rightarrow \left(5x-1\right)\left(7x-19\right)=0$
$\Rightarrow x=\frac{1}{5},\frac{19}{7}$
But $|x|<1,\therefore x=1/5$