So here is the Question :-
In the figure, $AB = 200$ is a diameter of the circle. Points $A$,$B$ are given the numbers $1,1$. The two semi-circles are bisected at $C,D$. Numbers $2,2$ are given $C,D$. Each, of the quarter circles are bisected and given the number $3$.
Now each arc is bisected and the point is given the number which is the sum of the $2$ numbers at the endpoints of the respective arc. This process is the continued till the sum of all numbers on the circle is at-least numerically equal to the product of $4$ times the area of the circle and $\frac{1}{3}$rd the radius. Find the least number of points on the circle (Take $\pi$ to be $3.14$) .
What I Tried :- You can actually calculate the product of $4\pi r^2$ and $\frac{r}{3}$ , which comes out to be :- $$\frac{4\pi r^3}{3} \rightarrow \frac{4*3.14*1000000}{3} \rightarrow 4186666$$
Here $4186666$ is the numerical part which is the sum of all the numbers on each arcs of the circle . The question is, how do you get this sum, because there is no specific pattern I can observe between these numbers on the arcs?
Can someone help me?
You can't see the forest because you're too focused on the trees. Add all the numbers up after each step -- 2, then 6, then 18, ... seems to triple each time.
When you make an iteration by bisecting arcs, the new sum counts each addend from the previous sum three times -- once from the earlier point, and once from each newly generated neighbor of that original point. For example, when you generate the 4 and 5 elements, each 3 from the previous set is also counted as part of the adjacent 4 and part of the adjacent 5.
So the sum is $2×3^n$ where $n$ is the number of iterations after starting with $A$ and $B$. In addition each bisecting doubles the count of points, as the number of arcs is also doubled. Proceed from there.