Which of the following sequences can be approximated by 4n^3? (1 point)
It is not clear what the given sequences are. Please provide the sequences to determine if they can be approximated by 4n^3.
To determine which of the given sequences can be approximated by the expression 4n^3, we need to compare the growth rate of each sequence to the growth rate of 4n^3.
The expression 4n^3 represents a cubic growth rate since the highest power of n is 3. Therefore, we are looking for sequences that also exhibit a cubic growth rate.
Let's examine each sequence:
1. 2n^2:
This expression represents a quadratic growth rate since the highest power of n is 2. It cannot be approximated by 4n^3.
2. 3n^3 + n:
This expression also represents a cubic growth rate since the highest power of n is 3. It can be approximated by 4n^3.
3. n! (n factorial):
This expression represents an increasing factorial growth rate. The factorial function grows much faster than any power of n, including cubic growth. Therefore, it cannot be approximated by 4n^3.
In summary, the sequence that can be approximated by 4n^3 is 3n^3 + n.
To determine which of the following sequences can be approximated by the expression 4n^3, we need to compare the given expression with each sequence.
The expression 4n^3 represents a cubic function, where the variable n is raised to the power of 3. This means that the value of the expression increases rapidly as n increases.
To compare this with the given sequences, we need to consider how the terms of the sequences grow as we increase n. Let's analyze each sequence:
1. Sequence 1: 2, 16, 54, 128, 250...
2. Sequence 2: 8, 64, 216, 512, 1000...
3. Sequence 3: 1, 8, 27, 64, 125...
Now, let's compare each term of the sequences with the expression 4n^3:
1. For Sequence 1, if we replace n with 1, we get 4(1)^3 = 4. However, the first term of Sequence 1 is 2, which is not equal to 4. So, Sequence 1 cannot be approximated by 4n^3.
2. For Sequence 2, if we replace n with 1, we get 4(1)^3 = 4. The first term of Sequence 2 is 8, which is equal to 4(1)^3. Similarly, if we replace n with 2, we get 4(2)^3 = 32, which matches the second term of Sequence 2. Thus, Sequence 2 can be approximated by 4n^3.
3. For Sequence 3, if we replace n with 1, we get 4(1)^3 = 4. The first term of Sequence 3 is 1, which is not equal to 4. So, Sequence 3 cannot be approximated by 4n^3.
Based on the analysis, Sequence 2 can be approximated by 4n^3.