Ratio of odd terms of a geometric progressions

Question

question:express the ratio of the sum of the squares of the odd number of terms of a GP to the sum of those terms as a polynomial of the common ratio of the GP

My try:I have first let a GP with $2n+1$ terms $a, ar, ar^2....ar^{2n}$ then odd terms are $a, ar^2, ar^2,..., ar^{2n}$ then the square of odd terms are $a^2, a^2r^4, a2r^8...a^2r^{4n}$

Then using the sum of gp formula and solving further I got

$a [\dfrac {r^{2 (n+1)}+1}{r^2+1}$

on dividing the above fraction I got $a[1-r^2+...-r^{2n-4}+r^{2n-2}]$ but the correct answer is $a[1-r+r^2....-r^{2n-1}+r^{2n}]$

But , I am not getting further, how to solve, please help

Answer 1

the sum of the squares of the odd number of terms of a GP

This should mean $$a^2+(ar)^2+(ar^2)^2+\cdots +(ar^{2n})^2$$

Hence, the ratio is given by $$\frac{a^2+(ar)^2+(ar^2)^2+\cdots +(ar^{2n})^2}{a+ar+ar^2+\cdots +ar^{2n}}$$ which can be written as $$\frac{a^2(1-r^{4n+2})}{1-r^2}/\left(\frac{a(1-r^{2n+1})}{1-r}\right)=\frac{a(1+r^{2n+1})}{1+r}=a(1-r+r^2-\cdots -r^{2n-1}+r^{2n})$$