1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4

a.15
b.15/2
c.-15
d.1/15
answer=a

2)Find the sum of the infinite geometric series n=1 at the bottom of the sigma notation 20(-1/4)n-1 on the right side
a.25
b.80/3
c.16
d.does not exist
answer=c

3)Find the sum of the infinite geometric series :4+3+9/4+:...
a.16/7
b.16
c.-12
d.does not exist
answer=c

4)Write Find the third iterate x3 o f(x)=x^2-4 for an initial value of x0=2
a.-4
b.4
c.12
d.-12
answer=c

5) 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
answer=c

We will be happy to critique your work, all of these are straightforward use of formula. We are not going to do the work for you.

1) To find the value of a1 in a geometric series, we can use the formula Sn = a1 * (1 - r^n) / (1 - r), where Sn is the sum of the series, a1 is the first term, r is the common ratio, and n is the number of terms.

Given: Sn = 300, r = -3, n = 4

Substituting the given values into the formula, we have:
300 = a1 * (1 - (-3)^4) / (1 - (-3))

Simplifying the equation:
300 = a1 * (1 - 81) / 4
300 = a1 * (-80) / 4
1200 = -80 * a1

Dividing both sides by -80:
a1 = 1200 / -80
a1 = -15

So, the value of a1 in the geometric series is -15.

2) To find the sum of an infinite geometric series, we can use the formula S = a / (1 - r), where S is the sum of the series, a is the first term, and r is the common ratio.

Given: a = 20, r = -1/4

Substituting the values into the formula, we have:
S = 20 / (1 - (-1/4))

Simplifying the equation:
S = 20 / (1 + 1/4)
S = 20 / (5/4)
S = 20 * (4/5)
S = 16

So, the sum of the infinite geometric series is 16.

3) To find the sum of an infinite geometric series, we can use the formula S = a / (1 - r), where S is the sum of the series, a is the first term, and r is the common ratio.

Given: a = 4, r = 3/4

Substituting the values into the formula, we have:
S = 4 / (1 - (3/4))

Simplifying the equation:
S = 4 / (1/4)
S = 4 * (4/1)
S = 16

So, the sum of the infinite geometric series is 16.

4) To find the third iterate of a function, we repeatedly apply the function to the initial value.

Given: f(x) = x^2 - 4, x0 = 2

Applying the function three times:
x1 = f(x0) = (2^2) - 4 = 0
x2 = f(x1) = (0^2) - 4 = -4
x3 = f(x2) = (-4^2) - 4 = -12

So, the third iterate of the function f(x) = x^2 - 4, with an initial value of x0 = 2, is -12.

5) To verify if n = 1 is a counterexample to the formula, we substitute n = 1 into the equation and see if it holds.

Given formula: 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = n * (2n + 1) / 3

Substituting n = 1 into the formula:
1^2 = 1 * (2*1 + 1) / 3
1 = 1 * 3 / 3
1 = 1

Since the equation holds true when n = 1, it is not a counterexample to the formula.

Therefore, the answer is not option c) n = 1.