Sum of the squares of the terms which are in infinite Geometric progression(GP)
Let us consider an infinite GP be \[a,ar,ar^2,ar^3,..................\]
here first term is a and common ratio is r
on squaring it we get \[a^2,a^2r^2,a^2r^4,a^2r^8,..................\]
on summing it up we get \[a^2+a^2r^2+a^2r^4+a^2r^8+..................\]
taking \[a^2\] common we get \[a^2(1+r^2+r^4+r^8+......)\] which is an infinite GP
sum of any infinite GP is given by \[\frac{x}{1-k}\]
where x is first term and k is common ratio
In this case x=a and \[k=r^2\] hence sum is given by \[a^2{\frac{1}{1-r^2}}\]
or sum =\[ \frac{a^2}{1-r^2}\]
Boundary condition r must lie between -1 to 1
Example- find the sum of the squares of \[1,\frac{1}{2},\frac{1}{4},\frac{1}{8}\]...........till infinity
here a =1 and r =\[ \frac{1}{2}\]
hence sum of the squares =\[\frac{1^2}{1-{\frac{1}{2}}^2}=\frac{4}{3}\]
