∴OA 1= 1

∵B 1A 1⊥x axis

The abscissa of point B 1 is 1, and point B 1 is on a straight line.

So y= root number 3.

So B 1=( 1, root number 3)

So A 1B 1= root number 3.

In Rt△A 1B 1O, by pythagorean theorem, we get

OB 1=2

∴sin∠ob 1a 1= 1/2

∴∠OB 1A 1=30

∴∠ob 1a 1=∠ob2a2=∠ob3a3=…=∠obnan=30

∫OA2 = ob 1 = 2，A2(2，0)

In Rt△OB2A2, OB2=2OA2=4.

∴OA3=4, A3 (4 4,0) The same is true.

OA4=8，…，0An=2n- 1，An(2n- 1，0)

∴OA5=25- 1= 16

∴A5( 16,0).

So the answer is: (16,0).

Adopt the best answer