Choose the THREE options that define sequences that do not converge.

Options
A P0 = 40, Pn+1 − Pn = 2.8 Pn (1− Pn/300) (n = 0,1,2, . . .)
B P0 = 100, Pn+1 − Pn = 0.7Pn (1− Pn/480) (n = 0,1,2, . . .)
C P0 = 250, Pn+1 − Pn = 2.4Pn (1− Pn/420) (n = 0,1,2, . . .)
D an = (5−3n)/(7n + 12) (n = 0,1,2, . . .)
E an =50 /(5(0.2))^n (n = 0,1,2, . . .)
F an =(8n^4 + 10n^2) / (4−3n^5) (n = 0,1,2, . . .)

To determine which options define sequences that do not converge, we need to examine the behavior of each sequence as n approaches infinity. If a sequence does not approach a specific value or become arbitrarily close to a value, it is considered to not converge.

Let's analyze each option:

Option A: P0 = 40, Pn+1 − Pn = 2.8 Pn (1− Pn/300) (n = 0,1,2, . . .)
To determine if this sequence converges, we can start by calculating the first few terms and observe the pattern. However, since the formula for calculating Pn+1 depends on Pn, it's not as straightforward to determine convergence. Thus, we need to employ a different approach.

By analyzing the recursive formula, we can try to investigate whether the sequence is bounded or unbounded. If the terms of the sequence grow larger and larger without any bound, then the sequence does not converge.

Let's assume Pn converges to a certain value L as n approaches infinity. We can rewrite the recursive formula as:
Pn+1 = 2.8Pn - (2.8Pn^2 / 300)
Pn+1 ≈ 2.8Pn

If Pn converges to L, then Pn+1 would also converge to L, since L does not depend on the value of n. However, this is not the case because Pn+1 is significantly smaller than 2.8Pn, indicating that the terms of the sequence grow without bound. Therefore, option A does not converge.

Option B: P0 = 100, Pn+1 − Pn = 0.7Pn (1− Pn/480) (n = 0,1,2, . . .)
Using the same reasoning as above, we can approximate the recursive formula as:
Pn+1 ≈ 0.7Pn

Similar to option A, if Pn converges to L, then Pn+1 would also converge to L. However, in this case, Pn+1 is significantly smaller than 0.7Pn, suggesting that the terms grow without bound. Therefore, option B does not converge.

Option C: P0 = 250, Pn+1 − Pn = 2.4Pn (1− Pn/420) (n = 0,1,2, . . .)
Once again, we can approximate the recursive formula as:
Pn+1 ≈ 2.4Pn

If Pn converges to L, then Pn+1 would also converge to L. However, as with options A and B, Pn+1 is much smaller than 2.4Pn, indicating that the terms of the sequence grow without bound. Therefore, option C does not converge.

Option D: an = (5−3n)/(7n + 12) (n=0,1,2,...)
To determine if this sequence converges, we can analyze the behavior of the terms as n approaches infinity. By examining the expression, we can see that the numerator, 5-3n, decreases as n increases, while the denominator, 7n+12, increases as n increases.

As n approaches infinity, the dominant term in the denominator, 7n, will grow much faster than the numerator, 5-3n. Therefore, the terms of the sequence will become arbitrarily close to zero. This indicates that the sequence converges to 0.

Option D converges.

Option E: an = 50 /(5(0.2))^n (n = 0,1,2, . . .)
In this sequence, we have an exponential term with a base less than 1, indicating that the terms of the sequence will decrease drastically as n increases.

As n approaches infinity, the exponent (0.2)^n will approach zero, causing the terms to grow without bound in the negative direction. Therefore, the sequence does not converge.

Option E does not converge.

Option F: an = (8n^4 + 10n^2) / (4−3n^5) (n = 0,1,2, . . .)
When analyzing this sequence, we observe that the denominator, 4-3n^5, can approach zero as n increases. If the denominator becomes zero at any point, it will result in an undefined term, which means the sequence does not converge.

To determine if the denominator reaches zero, we can solve the equation 4-3n^5 = 0.
n^5 = 4/3
n = (4/3)^(1/5)

Since the exponent n can take fractional values less than 1, option F does not converge.

In summary, the three options that define sequences that do not converge are:
A) P0 = 40, Pn+1 − Pn = 2.8 Pn (1− Pn/300) (n = 0,1,2, . . .)
B) P0 = 100, Pn+1 − Pn = 0.7Pn (1− Pn/480) (n = 0,1,2, . . .)
C) P0 = 250, Pn+1 − Pn = 2.4Pn (1− Pn/420) (n = 0,1,2, . . .)