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)
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
= 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
3. (15a^4 b^2) where a =n and b = -2p
=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:
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
3n^2 + 11n + 8 = 3n^2 + 11n + 8
Lo and behold, it is true
So its D? I dont 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
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