# math

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.

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3. 17
1. 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

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posted by Reiny
2. Im homeschooled its algebra II

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posted by Jon
3. 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)?

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posted by Damon
4. 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

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posted by Jon
5. 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

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posted by Damon
6. So its D? I don't understand

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posted by Jon
7. 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

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posted by Damon
8. no im not its a multiple choice
question

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posted by Jon

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