Question

If the sum to infinite of the series $1+4x+7x^2+10x^3+...$ is $\frac{35}{16}$, then find $x$.
(Note : $|x|<1$)

• A. $\frac{1}{5}$
• B. $\frac{-1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{5}$

Let $S_{\infty }=1+4x+7x^2+10x^3+...\infty$ $...\left(1\right)$
Now, multiply by $x$ throughout in eqution $\left(1\right)$; we get
$x{S}_{\infty }=x+4x^2+7x^3+10x^4+...\infty$ $...\left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$; we get
$S_{\infty }-x{S}_{\infty }=(1+4x+7x^2+10x^3+...\infty )-(x+4x^2+7x^3+10x^4+...\infty )$
$\Rightarrow (1-x)S_{\infty }=1+4x+7x^2+10x^3+...\infty -x-4x^2-7x^3-10x^4-...\infty$
$\Rightarrow (1-x)S_{\infty }=1+3x+3x^2+3x^3+...\infty$
Notice that the series $3x+3x^2+3x^3+...\infty$ is geometric series with the first term $a=3x$ and the common ratio $r=x$.
Now, use the formula for the sum of an infinite geometric series.
$\Rightarrow (1-x)S_{\infty }=1+\frac{3x}{(1-x)}$, for$\left|x\right|<1$
$\Rightarrow (1-x)S_{\infty }=\frac{1+2x}{(1-x)}$, for$\left|x\right|<1$
Given that, $S_{\infty }=\frac{35}{16}$, substitute for $S_{\infty }$ in the above equation; we get
$(1-x)\frac{35}{16}=\frac{1+2x}{(1-x)}$
$\Rightarrow 35{(1-x)}^2=16(1+2x)$
$\Rightarrow 35(1-2x+x^2)=16+32x$
$\Rightarrow 35x^2-102x+19=0$
$\Rightarrow (7x-19)(5x-1)=0$
$\Rightarrow x=\frac{19}{7}$ or $x=\frac{1}{5}$
But $x\ne \frac{19}{7}$, because for infinity series, $\left|x\right|<1$.
Therefore, $x=\frac{1}{5}$.