Neglecting the first term

We have $1= r + r^{2} + r^{3} +.... = r + r( r+ r^{2} + r^{3} +...) = r+ r \times 1 = 2r$ ,

So $r= \frac{1}{2}$ . Therefore

$r^2 + r^3 + ... = 1-r = \frac{1}{2}$ .

If $\large r + r^2 + r^3 + \cdots = 1$ , then $\large r^2 + r^3 + r^4 + \cdots = r$ (by multiplying both sides by $r$ )

Then $\large r+r = 1 \implies r=\frac{1}{2}$

Now we see $\large \sum_{n=2}^{\infty} r^n =\frac{a_1}{1-r}$ ( Infinte sum of geometric progression where $a_1=r^2$ and $r=\frac{1}{2}$ ).

Thus the sum will be equal to $\Large\frac{\Large\frac{1}{4}}{\Large\frac{1}{2}}= \frac{1}{2}$

Thats quite cool. You first add the firstG.P you get value of r = 1/2. Now the GP given below is =(GP given above) - r = 1-1/2 = 1/2 Ans