Sum of the cubes of the terms which are in infinite geometric progression is given by \[\frac{a^3}{1-r^3}\]
proof
let us consider a infinite geometric series \[a , ar ,ar^2,ar^3,.............\infty \]
sum of the cubes can be written as \[a^3+a^3r^3+a^3r^6+.....\infty \]
taking \[a^3 \] common it can written as \[a^3 (1+r^3+r^6+....................+\infty )\]
which can further be written as as sum of infinite GP with first term =1
and common ratio=r^3
hence the sum =\[a^3 (\frac{1}{1-r^3})\]
=\[(\frac{a^3}{1-r^3})\]
Example - Find the cubes of the series \[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......\infty \]
solution
sum of the cubes of the term in infinite geometric progression can be given by \[\frac{a^3}{1-r^3}\]
here a=1 and r=\[\frac{1}{2}\]
hence sum=\[\frac{1^3}{1-(1/2)^3}\]
=\[\frac{8}{7}\]
