math
posted by Jon on .
1)Find the third iterate x3 of f(x)=x24 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:(wx)5
This ones long so I chose w55w4x+10w3x310w2x4+5wx4x5
3)Use the binomial Theorem to find the third term in the expression of:(n2p)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+...+(2n1)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+...+(3n1)=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
t_{r+1} = (6 choose r)a^(6r)b^r
so for (n2p)^6
t_{3} = (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 = 44 = 0
X2 = 04 =4
X3 = 164 = 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^210w^2x^3+5wx^4x^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)=x24 for an initial value of x0=2
thats how it is in my book
2)the choices are:
A)w^54w^4x+6w^3x^26w^2x^3+4wx^4x^5
B)w^5+5w^w10^3x^2+10w^2x^35wx^4+x^5
C)w^5+4w^4x6w^3x^2+6w^2x^34wx^4+x
D)w55w4x+10w3x310w2x4+5wx4x5 
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