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
recheck 1,3
I don't understand 2
do you know about sigma notation?
1) To find a1 in a geometric series, we can use the formula for the nth term in a geometric series: an = a1 * r^(n-1), where a1 is the first term, r is the common ratio, and n is the number of terms.
Given that Sn = 300, r = -3, and n = 4, we can use the formula for the sum of a geometric series: Sn = a1 * (1 - r^n) / (1 - r). Plugging in the values, we have:
300 = a1 * (1 - (-3)^4) / (1 - (-3))
Simplifying the equation, we get:
300 = a1 * (1 - 81) / 4
300 = a1 * (-80) / 4
Multiplying both sides by 4 and dividing by -80, we get:
a1 = 300 * (-1) / 80 = -3.75
So, the first term, a1, is -3.75. However, the given answer choices do not include this value. Therefore, none of the given answer choices are correct.
2) To find the sum of an infinite geometric series, we can use the formula: S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio.
Given the series n=1 on the bottom and 20(-1/4)^(n-1) on the right side, we can find the first term by plugging in n = 1:
a = 20 * (-1/4)^(1-1) = 20 * (-1/4)^0 = 20 * 1 = 20
Next, we can find the common ratio by plugging in n = 2 (as the series starts from n = 1):
r = 20 * (-1/4)^(2-1) = 20 * (-1/4)^1 = 20 * (-1/4) = -20/4 = -5
Now we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r) = 20 / (1 - (-5)) = 20 / (1 + 5) = 20 / 6 = 10/3
So, the sum of the infinite geometric series is 10/3. However, none of the given answer choices are correct.
3) To find the sum of an infinite geometric series, we can use the formula: S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio.
Looking at the given series: 4 + 3 + 9/4 + ..., we can see that the first term, a, is 4.
To find the common ratio, we can divide each term by its preceding term:
3 / 4 = 0.75
(9/4) / 3 = 9/12 = 0.75
As both ratios are equal to 0.75, we can conclude that the common ratio, r, is 0.75.
Using the formula for the sum of an infinite geometric series:
S = a / (1 - r) = 4 / (1 - 0.75) = 4 / 0.25 = 16
So, the sum of the infinite geometric series is 16. Hence, the correct answer is (c).
4) To find the third iterate (x3) of the function f(x) = x^2 - 4 for an initial value of x0 = 2, we need to apply the function three times to the initial value.
First iteration: f(x0) = f(2) = 2^2 - 4 = 4 - 4 = 0
Second iteration: f(f(x0)) = f(0) = 0^2 - 4 = -4
Third iteration: f(f(f(x0))) = f(-4) = (-4)^2 - 4 = 16 - 4 = 12
So, the third iterate (x3) of the function is 12. Therefore, the correct answer is (c).
5) To check which option is not a counterexample to the formula 1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = n(2n + 1) / 3, we need to substitute the values of n from each option into the formula and check if the equation holds true.
Let's substitute the values:
a) n = 3:
1^2 + 3^2 + 5^2 = 3(2*3 + 1) / 3
1 + 9 + 25 ≠ 3(6 + 1) / 3
35 ≠ 21
This option is a counterexample.
b) n = 2:
1^2 + 3^2 = 2(2*2 + 1) / 3
1 + 9 ≠ 2(4 + 1) / 3
10 ≠ 10/3
This option is not a counterexample.
c) n = 1:
1^2 ≠ 1(2*1 + 1) / 3
1 ≠ 3/3
This option is a counterexample.
d) n = 4:
1^2 + 3^2 + 5^2 + 7^2 = 4(2*4 + 1) / 3
1 + 9 + 25 + 49 ≠ 4(8 + 1) / 3
84 ≠ 36
This option is a counterexample.
Therefore, the option that is NOT a counterexample to the formula is (b) n = 2.