Question

If sum to infinity of the series $3-5r+7r^2-9r^3+....$ is $\frac{14}{9}$, find r.

Answer 1

$S=3-5r+7r^2-9r^3......\infty$

$Sr=3r-5r^2+7r^3-9r^4......\infty$

$(1+r)S=3-2r+2r^2-2r^3....\infty$

$(1+r)S=3-2(r-r^2+r^3.....\infty )$

$S=\frac{3}{1+r}-\frac{2r}{(1+r{)}^2}$

$\frac{14}{9}=\frac{3+r}{(1+r{)}^2}$

$14(1+2r+r^2)=27+9r$

$14r^2+19r-13=0$

$r=0.5$ and $r=-1.857$