
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