1)Find the third iterate x3 of f(x)=x2-4 for an initial value of x0=2

A)-4
B)4
C)12
D)-12
I chose C

2)Use Pascal's triangle to expand:(w-x)5
This ones long so I chose w5-5w4x+10w3x3-10w2x4+5wx4-x5

3)Use the binomial Theorem to find the third term in the expression of:(n-2p)6
A)60n4p2
B)120n3p3
C)-12n5p
D)-160n2p4
I chose C

4)Which is not a counterexample to the formula:1^2+3^2+5^2+...+(2n-1)2=n(2n+1/3
A)n=3
B)n=2
C)n=1
D)n=4
I chose C

5)In an induction proof of the statement 4+7+10+...+(3n-1)=n(3n+5)/2
the first step is to show that the statement is true for some integers n.
Note:3(1)+1=1[3(1)+5]/2 is true. Select the steps required to complete the proof.
A)Show that the statement is true for any real number k. Show that the statement is true for k+1.
B)Assume that the statement is true for some positive integer k. Show that the statement is true for k+1.
C)Show that the statement is true for some positive integers k. Give a counterexample.
D)Assume thst the statement is true for some positive integers k+1. Show that the staement is true for k.
I don't know

Thank you SO much Reiny for helping me. I'm not looking for anyone to give me the answer because Ive done the work and what ever I answered is my BEST answer and I can't afford to get them wrong so thanks for all your help from the bottom of my heart.

for #1, I don't know what you mean by "third iterate", what course and grade uses that terminology?

for #2, for every term the total of the exponents should be 5, your two middle terms are incorrect, they should be
10(w^3)(x^2) and -10(x^2)(x^3)

#3
in the binomial expansion for (a+b)^n
tr+1 = (6 choose r)a^(6-r)b^r

so for (n-2p)^6

t3 = (6 choose 2)n^4(-2p)^2
= 15n^4(4p^2)
= 60n^4p^2 which is A

Im homeschooled its algebra II

1. do you mean Xn+1 = Xn^2 - 4 ?

X0 = 2
X1 = 4-4 = 0
X2 = 0-4 =-4
X3 = 16-4 = 12

2. row five 1 5 10 10 5 1
wherever you have an odd power of x, you will have a minus sign with a =w and b = -x
w^5 -5w^4x^1+10w^3x^2-10w^2x^3+5wx^4-x^5

3. (15a^4 b^2) where a =n and b = -2p

(15)n^4(-2p)^2
=60 n^4 p^2

4.Yes c, but PLEASE use parentheses carefully and use ^ for to the power! I am spending all my time figuring out what you mean.

5. I do not understand. Are you sure it is not (3n+1)?

1)Find the third iterate x3 of f(x)=x2-4 for an initial value of x0=2

thats how it is in my book

2)the choices are:
A)w^5-4w^4x+6w^3x^2-6w^2x^3+4wx^4-x^5
B)w^5+5w^w-10^3x^2+10w^2x^3-5wx^4+x^5
C)w^5+4w^4x-6w^3x^2+6w^2x^3-4wx^4+x
D)w5-5w4x+10w3x3-10w2x4+5wx4-x5

5.Assuming this works, find out

Using (3n+1) which is 3(n+1)+1 at n+1 which is 3n+4

Sum at(n+1) = sum at n + 3(n+1)+1
(n+1)(3(n+1)+5)/2 =n(3n+5)/2 + 3n+4 multiply by two and simplify
(n+1)(3n+8) = 3n^2 + 5n +6n +8
or
3n^2 + 11n + 8 = 3n^2 + 11n + 8
Lo and behold, it is true

So its D? I don't understand

In number 5 you are supposed to do A through D

If you can do one I think you can do them all. I did it for, n and n+1

no im not its a multiple choice

question

You're welcome! I'm glad I could assist you. It's great that you've put in the effort and have come up with your best answers. For the questions you're unsure about, I can help guide you through the process of finding the correct answer. Let's go through them:

1) Find the third iterate x3 of f(x) = x^2 - 4 for an initial value of x0 = 2.
To find the third iterate, we need to apply the function f(x) three times.

- First iteration: f(2) = (2)^2 - 4 = 0.
- Second iteration: f(0) = (0)^2 - 4 = -4.
- Third iteration: f(-4) = (-4)^2 - 4 = 12.

Therefore, the third iterate x3 is equal to 12. So the correct answer is C) 12.

2) Use Pascal's triangle to expand: (w - x)^5
Pascal's triangle can be used to expand binomial expressions. The coefficients for each term are the corresponding numbers in the row of Pascal's triangle. The exponents on w decrease while the exponents on x increase.

The fifth row of Pascal's triangle is 1 5 10 10 5 1.

Expanding (w - x)^5 using these coefficients:
(w - x)^5 = (w^5) - (5w^4x) + (10w^3x^2) - (10w^2x^3) + (5wx^4) - (x^5)

So, the correct expansion is (w^5) - (5w^4x) + (10w^3x^2) - (10w^2x^3) + (5wx^4) - (x^5). Therefore, your answer is correct.

3) Use the binomial theorem to find the third term in the expansion of (n - 2p)^6.
The binomial theorem states that the kth term in the expansion of (a + b)^n is given by the formula: C(n, k) * a^(n-k) * b^k, where C(n, k) is the binomial coefficient, also known as "n choose k".

In this case, the third term corresponds to k = 2.
Using the formula: C(6, 2) * (n)^(6-2) * (-2p)^2

C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15.

Therefore, the third term is 15 * n^4 * 4p^2 = 60n^4p^2. So the correct answer is A) 60n^4p^2.

4) Which is not a counterexample to the formula: 1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = n(2n + 1) / 3.
To determine the counterexample, we need to find a value of n that makes the formula false.

- For n = 3: 1^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35 ≠ 3(2(3) + 1) / 3 = 21.
- For n = 2: 1^2 + 3^2 = 1 + 9 = 10 ≠ 2(2(2) + 1) / 3 = 10/3.
- For n = 1: 1^2 = 1 ≠ 1(2(1) + 1) / 3 = 1 / 3.
- For n = 4: 1^2 + 3^2 + 5^2 + 7^2 = 1 + 9 + 25 + 49 = 84 ≠ 4(2(4) + 1) / 3 = 40/3.

Therefore, the value of n that is not a counterexample is C) n = 1.

5) In an induction proof of the statement 4 + 7 + 10 + ... + (3n - 1) = n(3n + 5) / 2, we're given the initial step that shows the statement is true for n = 1: 3(1) - 1 = 1[3(1) + 5] / 2.

To complete the proof, we need to follow the steps for mathematical induction, which are as follows:
B) Assume that the statement is true for some positive integer k. Show that the statement is true for k + 1.

Using this assumption, let's substitute k into the equation and then show that it holds true for k + 1.

Let's assume the statement is true for k:
4 + 7 + 10 + ... + (3k - 1) = k(3k + 5) / 2

Now, we need to show that the statement is true for k + 1:
4 + 7 + 10 + ... + (3k - 1) + (3(k + 1) - 1) = (k + 1)(3(k + 1) + 5) / 2

From here, continue expanding and simplifying both sides of the equation until they match.

Therefore, the correct next step is B) Assume that the statement is true for some positive integer k. Show that the statement is true for k + 1.

Keep up the good work! If you have any more questions, feel free to ask.