Question

If the sum to infinity of the series $2+5x+8x^2+11x^3+......\infty$ is $\frac{20}{9}$ then find x where $\left|x\right|<1$

Answer 1

Given series is an AGP. Follow the same procedure
Multiply the series with the common ration and the subtract it with original series
Let $s_{\infty }=2+5x+8x^2+11x^3+.....\infty$ .....(1)
Multiplying both side of equation 1 by x
$x{s}_{\infty }=2x+5x^2+8x^3+.....\infty$ .....(2)
Subtracting equation 2 from equation 1 we get
$(1-x)s_{\infty }=2+3x+x^2+x^3+......\infty$

$(1-x)s_{\infty }=2+\frac{3x}{1-x}=\frac{2-2x+3x}{1-x}=\frac{1+x}{1-x}$
$(1-x)s_{\infty }=\frac{1+x}{(1-x{)}^2}=\frac{20}{9}$
$9+9x=20+20x^2-40x$
$20x^2-5x-44x+11=0$
$5x(4x-1)-11(4x-1)=0$
$(4x-1)(5x-11)$
$x=\frac{1}{4},\frac{11}{5}$
Since $\left|x\right|<1$
$x=\frac{1}{4}$