91Ó°ÊÓ

The following is the required BASIC Program-

Let variable P which locate the position of points. Bee starting from point $P_0$which flies to point $P_1$and lands where 0 and 1 are the coordinates of points. Then bee returns half of the way to $P_0$, landing at $P_2$. From $P_2$, the bee returns half of the way to $P_1$, landing on $P_3$and so forth. Thus, through BASIC program compute and print the bee’s location after 2 to 10 trips.

Let P and N are two variables where P is used for the position of the bee and N is for the number of position changes by the bee and $P\left(N\right)$is the distance between the initial positions of the bee to the actual position of the bee.

10 dim $P\left(50\right)$ //$P\left(50\right)$ shows the bee position can change 50 times.

20 LET $P\left(0\right)=0$

30 LET $P\left(1\right)=1$

40 For $N=2$to 10

50 LET $P\left(N\right)=P\left(N−1\right)+\left(−1/2\right)↑\left(N−1\right)$

60 PRINT $N,P\left(N\right)$

70 NEXT N

80 END

Sample Output:

2 0.5

3 0.75

4 0.625

5 0.6875

6 0.65625

7 0.671875

8 0.6640625

9 0.66796875

10 0.66601562

The result is the same as the program 1.

When executed program asked for values which are going to be executed in loops. N assigned value $2,3,4,5,6,7,8,9,10$when loop runs. And P(N) as its result that is $0.5,0.75,0.625,0.6875,0.65625,0.671875,0.6640625,0.66796875,0.66601562$.

This result is as same as program 1.

Consider the equation given below that is as like formula;

$P\left(N\right)=P\left(N−1\right)+\left(−1/2\right)↑\left(N−1\right)$

$1−\frac{1}{2}+\frac{1}{4}−\frac{1}{8}+\mathrm{.....}+{\left(−\frac{1}{2}\right)}^{n−1}$

Here,

For $n=2$,$1−\frac{1}{2}=0.5$

For $n=3$, $1−\frac{1}{2}+\frac{1}{4}=0.75$and so on.

Thus, each term series reflect the bee’s return half of the way from $P_{n−1}$to $P_{n−2}$.

This equation shows as same as the program been written above.