# 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

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

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

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

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

## 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.