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
1) To find the value of a1 in a geometric series, we can use the formula for the nth term of a geometric series:
an = a1 * r^(n-1)
In this case, we are given Sn (the sum of the first n terms), r (the common ratio), and n (the number of terms). We need to find a1.
To find a1, we can rearrange the formula for Sn:
Sn = a1 * (1 - r^n) / (1 - r)
Plugging in the given values, we have:
300 = a1 * (1 - (-3)^4) / (1 - (-3))
300 = a1 * (1 - 81) / (1 + 3)
300 = a1 * (-80) / 4
To solve for a1, we multiply both sides by 4 and divide by -80:
a1 = -15
Therefore, the answer is (c) -15.
2) To find the sum of an infinite geometric series, we can use the formula:
Sinf = a / (1 - r)
In this case, we are given a = 20 and r = -1/4.
Plugging in the values, we have:
Sinf = 20 / (1 - (-1/4))
Sinf = 20 / (1 + 1/4)
Sinf = 20 / (5/4)
Sinf = 20 * (4/5)
Sinf = 16
Therefore, the answer is (c) 16.
3) To find the sum of an infinite geometric series, we can use the formula for the sum:
Sinf = a / (1 - r)
In this case, we are given a = 4 and r = 3/4.
Plugging in the values, we have:
Sinf = 4 / (1 - 3/4)
Sinf = 4 / (1/4)
Sinf = 4 * (4/1)
Sinf = 16
Therefore, the answer is (c) 16.
4) To find the third iterate of a function f(x), we need to repeatedly apply the function three times starting with an initial value of x0.
Given the function f(x) = x^2 - 4 and x0 = 2, we can find x1, x2, and x3 as follows:
x1 = f(x0) = (2)^2 - 4 = 4 - 4 = 0
x2 = f(x1) = (0)^2 - 4 = -4
x3 = f(x2) = (-4)^2 - 4 = 12
Therefore, the answer is (c) 12.
5) To check which option is a counterexample to the given formula, we can substitute the values of n into the formula and see if it is true.
The given formula is 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = n(2n+1)/3
For option (c) n = 1:
1^2 = 1(2+1)/3
1 = 3/3
1 = 1
Since the equation is true for n = 1, option (c) is not a counterexample.
To check if any other options are counterexamples, we can substitute the values of n into the formula and see if they hold true. If any option does not satisfy the formula, then it is a counterexample.
Therefore, the answer is (c) n = 1.
1) To find the first term (a1) in the geometric series, we 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, and n = 4, we can plug these values into the formula and solve for a1:
300 = a1 * (1 - (-3)^4)/(1 - (-3))
300 = a1 * (1 - 81)/4
300 = a1 * (-80/4)
300 = a1 * -20
a1 = 300 / -20
a1 = -15
Therefore, the correct answer is c. -15.
2) The given geometric series can be represented as: Σ (20 * (-1/4)^(n-1)), where n is the iteration number.
To find the sum of the infinite series, we use the formula S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio.
Given a = 20 and r = -1/4, we can plug these values into the formula and solve for S:
S = 20 / (1 - (-1/4))
S = 20 / (1 + 1/4)
S = 20 / (5/4)
S = 20 * (4/5)
S = 80/5
S = 16
Therefore, the correct answer is c. 16.
3) To find the sum of the infinite geometric series, we use the formula S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio.
Given a = 4 and r = 3/4, we can plug these values into the formula and solve for S:
S = 4 / (1 - (3/4))
S = 4 / (1/4)
S = 4 * (4/1)
S = 16
Therefore, the correct answer is b. 16.
4) The iterates of a function f(x) represent the repeated application of the function to an initial value. In this case, we need to find the third iterate of the function f(x) = x^2 - 4, starting with an initial value of x0 = 2.
1st iterate: f(x0) = f(2) = (2)^2 - 4 = 4 - 4 = 0
2nd iterate: f(f(x0)) = f(0) = (0)^2 - 4 = -4
3rd iterate: f(f(f(x0))) = f(-4) = (-4)^2 - 4 = 16 - 4 = 12
Therefore, the correct answer is c. 12.
5) To find a counterexample to the given formula 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = n(2n+1)/3, we need to find a value of n that does not satisfy the equation.
Let's check each given option:
a) n = 3:
1^2 + 3^2 + 5^2 = 9 + 9 + 25 = 43
3(2(3) + 1)/3 = 3(6 + 1)/3 = 3(7)/3 = 7
Since 43 does not equal 7, option a is NOT a counterexample.
b) n = 2:
1^2 + 3^2 = 1 + 9 = 10
2(2(2) + 1)/3 = 2(4 + 1)/3 = 2(5)/3 = 10/3
Since 10 does not equal 10/3, option b is NOT a counterexample.
c) n = 1:
1^2 = 1
1(2(1) + 1)/3 = 1(2 + 1)/3 = 1(3)/3 = 1
Since 1 equals 1, option c is a counterexample.
d) n = 4:
1^2 + 3^2 + 5^2 + 7^2 = 1 + 9 + 25 + 49 = 84
4(2(4) + 1)/3 = 4(8 + 1)/3 = 4(9)/3 = 36/3 = 12
Since 84 does not equal 12, option d is NOT a counterexample.
Therefore, the correct answer is c. n = 1.