The nth term of sequence P is an + b

These are 14 questions and 14 answers.

Since this exceeds the limit and I had to delete the last questions and I copied all the answers to a file that is attache. See the attachment with all the answers.

1) Question 1. Find the first six terms of the sequence: a1 = -6, an = 4 • an-1

option D) -6, -24, -96, -384, -1536, -6144

Explanation:

A(1) = - 6

A(n) = 4 * A(n-1)

n                 A(n)

1                 - 6

2                  4 * (-6) = - 24

3                  4 * (-24) = - 96

4                  4 * (-96) = - 384

5                  4 * (-384) = - 1536

6                  4 * ( -1536) = -6144

So, the first six terms are: -6, - 24, - 96, - 384, - 1536, - 6144.

2) Question 2: Find an equation for the nth term of the arithmetic sequence.
-15, -6, 3, 12, ...

option D) - 15 + 9(n - 1)

Explanation:

1. find the difference between the consecutive terms:

-6 - (-15) = -6 + 15 = 9
3 - (-6) = 3 + 6 = 9
12 - 3 = 9

So, the difference is 9, and you can find any term adding 9 to the previous.

2. Since the first term is - 15, you have:

First term, A1 = - 15 + 9(0) = - 15
Second term, A2 =  - 15 + 9(1) = - 6
Third term, A3 = -15 + 9(2) = - 15 + 18 = 3
Fourth term, A4 = - 15 + 9(3) = - 15 + 27 = 12

3. So, the general formula is An = - 15 + 9 (n - 1), which is the option D)

3) Question 3. Find an equation for the nth term of the arithmetic sequence A14 = - 33, A15 = 9.

option B) An = - 579 + 42(n - 1)

Explanation:

1) Find the difference: 9 - (-33) = 9 + 33 = 42

2) A15 = A1 + 42 * (15 - 1)

=> A1 = A15 - 42(15 - 1)

A1 = A15 - 42(14)

A1 = 9 - 588 = - 579

Therefore, the formula es An = - 579 + 42(n - 1)

4) Question 4. Determine whether the sequence converges or diverges. If it converges, give the limit.

48, 8, 4/3, 2/9, ...

the sequence converges to 288/5

Explanation:

That is a geometric sequence.

The ratio is 1/6: 8/48 = 1/6; (4/3) / 8 = 4/24 = 1/6; (2/9)/(4/3) = 6/36 = 1/6.

The convergence criterium is that if |ratio| < 1 then the series, this is the sum of all the terms, converge to: A1 / (1 - ratio)

Then the limit 48 / (1 - 1/6) = 48 / (5/6) = 48*6 / 5 = 288/5

5) Question 5. Find an equation for the nth term of the sequence.

-3, -12, - 48, -192

- 3 * (4)^(n-1)

Explanation: clearly any term (from the second) is the previous term multiplied by 4.

The first term is  -3
The second term is -3(4) = - 12
The third term is -3(4)(4)= - 48
The fourth term is - 3 (4)(4)(4) = - 192

So, the general formula for the nth term is -3 * 4^ (n-1)

6) Question 6. Find an equation for the nth term of a geometric sequence where the second and fifth terms are -21 and 567, respectively.

An =7 * (-3)^(n-1)

Explanation:

1) The fith term is the second term * (ratio)^3: A5 = A3 * (r)^3

2)  A5 = 567, A2 = - 21 => r^3 = A5 / A2 = - 567 / 21 = - 27

=> r = ∛(-27) = - 3

3) So the first term is A1 = A2 / r = -21 / -3 = 7

4) The general formula is

An =7 * (-3)^(n-1)

7) Question 7. Write the sum using summation notation, assuming the suggested pattern continues.

5 - 15 + 45 - 135 + ...

option B) summation of five times negative three to the power of n from n equals zero to infinity

Explanation:

5 = 5
-15 = 5 (-3)
45 = 5(-3)^2
-135 = 5(-3)^3

=> 5 + 5(-3) + 5(-3)^2 + 5(-3)^3+

Using the summation notation that is:

∞
∑ (5)(-3)^n
n=0

Which means summation of five times negative three to the power of n from n equals zero to infinity

8) Question 8. Write the sum using summation notation, assuming the suggested pattern continues.
-9 - 3 + 3 + 9 + ... + 81

option A) summation of the quantity negative nine plus six n from n equals zero to fifteen

Explanation:

Find the difference:

-3 - (-9) = - 3 + 9 = 6
3 - (-3) = 3 + 3 = 6
9 - 3 = 6

First term: - 9
Second term: - 9 + 6(1)
Third term: - 9 + 6(2)

nth term = - 9 + (n -1)

Summation = [- 9] + [- 9 + 6(1) + [-9 + 6(2)] + [-9 + 6(3) ]+ [-9 + 6(15) ]

Using summation notation:

15
∑ [-9 + 6n]
n=0

which means summation of the quantity negative nine plus six n from n equals zero to fifteen.

9) Question 9. Write the sum using summation notation, assuming the suggested pattern continues.


64 + 81 + 100 + 121 + ... + n2 + ...

A) summation of n squared from n equals eight to infinity

Explanation:

64 = 8^2

81 = 9^2

100 = 10^2

121 = 11^2

n^2

=>

∞

∑ n^2

n=8

which means summation of n squared from n equals eight to infinity

10) Question 10. Find the sum of the arithmetic sequence.
17, 19, 21, 23, ..., 35

260

Explanation:

The difference is 2:

The sum is: 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35.

You can use the formula for the sum of an arithmetic sequence:

(A1 + An) * n / 2 = (17 + 35)*10/2 = 260

11) Question 11. Find the sum of the geometric sequence. 

1, 1/2, 1/4, 1/8, 1/16

option D) 31/16

Explanation:

You can either sum the 5 terms or use the formula for the partial sum of a geometric sequence.

The formula is: Sum = A * ( 1 - r^n) / (1 - r)

Here A = 1, r = 1/2, and n = 5 => Sum = 1 * (1 - (1/2)^5 ) / (1 - 1/2) =

= [ 1 - 1/32] / [1/2] = [31/32] / [1/2] = 31 / 16

A. We start by writting down the general formula:
a(1) = 10
a(2) = a(1) - 3  
a(3) = a(2) - 3
a(n) = a(n-1) -3

Now we find first ten terms:
10, 7, 4 ,1, -2, -5, -8, -11, -14, -17

b. We start by writting down the general formula:
a(1) = 1
a(2) = 1 + 2
a(3) = 1 + 2 + 3
a(n) = 1 + 2 + ... + n

Now we find first ten terms:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55

c. We start by writting down the general formula:
a(n) = 3n -2n = n

Now we find first ten terms:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

d. We start by writting down the general formula:


Now we find first ten terms:


e. We start by writting down the general formula:
a(1) = 1
a(2) = 5
a(3) = a(1) + a(2)
a(4) = a(2) + a(3)
a(n) = a(n-2) + a(n-1)

Now we find first ten terms:
1, 5, 6, 11, 17, 28, 45, 73, 118, 191

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A. We start by writting down the general formula:
a(1) = 10
a(2) = a(1) - 3  
a(3) = a(2) - 3
a(n) = a(n-1) -3

Now we find first ten terms:
10, 7, 4 ,1, -2, -5, -8, -11, -14, -17

b. We start by writting down the general formula:
a(1) = 1
a(2) = 1 + 2
a(3) = 1 + 2 + 3
a(n) = 1 + 2 + ... + n

Now we find first ten terms:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55

c. We start by writting down the general formula:
a(n) = 3n -2n = n

Now we find first ten terms:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

d. We start by writting down the general formula:


Now we find first ten terms:


e. We start by writting down the general formula:
a(1) = 1
a(2) = 5
a(3) = a(1) + a(2)
a(4) = a(2) + a(3)
a(n) = a(n-2) + a(n-1)

Now we find first ten terms:
1, 5, 6, 11, 17, 28, 45, 73, 118, 191

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1. A. According to the expression a_n=4*a_n-1, each term after a1 is four times the previous term. The first term is -7 as given, 2nd term should be -7*4=-28, 3rd term is -28*4=-112, ... A is the correct answer. 

2. B. The sequence is -13, -8, -3, 2... It's obvious that each term is equal to the previous term plus 5. This is an arithmetic sequence with initial term -13 and common difference 5. We know a1=-13, so a_n=-13+5*(n-1). The answer is B.

3. A. We are given a15=-53, a16=-5. The common difference of the arithmetic sequence is -5-(-53)=48. The formula for a_n term is a1+48*(n-1). We know that a15=-13; plug in n=15, a15=-53=a1+48*(15-1), a1=-725. So a_n=-725+48*(n-1).

4. Diverge. We are given a few terms, 11, 44, 176, 704... Observe that each term is four times the previous one. 11*4=44, 44*4=176, 176*4=704... This is a geometric series with common ratio>1. You can keep multiplying by 4 and the series goes to infinity, so it diverges.

5. D. We have -4, -16, -64, -256... Same as above, each term is four times the previous one. The initial term is a1=-4. The common ratio d=4. So a_n=a1*d^(n-1)=-4*4^(n-1)=-4^n. (D).

6. The answer is A. a2=-2, a5=16. Suppose the common ratio is D. a_n=a1*d^(n-1). a2=a1*d; a5=a1*d^4. Plug in a2 and a5: -2=a1*d, 16=a1*d^4. 16/-2=d^3=-8, d=-2, a1=1. So a_n=1*(-2)^(n-1).

7. B. We are given the sequence 4, -24, 144,... Each term is -6 times the previous one. The first term a0=4, the n^th term a_(n-1) is a1*d^n=4*(-6)^n. To express the sum, we simply have to use the sigma notation and sum 4*(-6)^n from n=0 to infinity. The answer is B.

8. D. We are given -3 + 6 + 15 + 24... 132. Each term is equal to the previous one plus 9. First term a0=-3, n^th term a_n-1 is -3+9*n. The last term is 132. 132 =-3+9n, n=15. So we have to sum -3+9n from n=0 to n=15.

9. B. 343 + 512 + 729 + 1000+...  343=7^3, 512=8^3, 729=9^3, 1000=10^3. This is a sequence of perfect cubes. Therefore, the sum is n^3 from n=7 to infinity. (The initial term is 343=7^3).

10. B. We are given 3, 5, 7, 9, ... 21. The common difference is 2. There are (21-3)/2+1=10  terms. The initial term a1=3, and last term is a10=21. The sum is (a1+a10)*10/2=(3+21)*10/2=120.

11. C. 4/3, 16/3, 64/3, 256/3, 1024/3.  Each term is four times the previous one. This is a geometric series with initial term a1=4/3 and common ratio r=4. 1024/3 is the 5th term of the sequence. So sum=a1*(1-r^n)/(1-r)=4/3*(1-4^5)/(1-4)=-4/9*-1023=1364/3.

12. B. 10,12,14,... This is an arithmetic sequence. a1=10, and common difference d=2. There are 20 terms (20 rows). a20=a1+d*(n-1)=10+2*(20-1)=48. So the sum S=(a1+an)*n/2=(10+48)*20/2=580.

13. 10 + 20 + 30 + ... + 10n = 5n(n + 1). When n=1, this expression is true, since 10=5*1*(1+1). Suppose when n=k, this statement is true, then when n=k+1, the left side is 10+...+10n+10(n+1), the right side is 5(n+1)(n+2). The left side adds 10(n+1) compared to the previous one. The right side adds 5(n+1)(n+2)-5n(n+1)=5(n+1)(n+2-n)=10(n+1). So the statement holds true.

14. The height at week 0 is a0=300 (initial height). Common difference is 4.2 (weekly increment). a_n=300+4.2n. At week n, the height of the tree is 300+4.2*n centimeters.

These are 14 questions and 14 answers.

Since this exceeds the limit and I had to delete the last questions and I copied all the answers to a file that is attache. See the attachment with all the answers.

1) Question 1. Find the first six terms of the sequence: a1 = -6, an = 4 • an-1

option D) -6, -24, -96, -384, -1536, -6144

Explanation:

A(1) = - 6

A(n) = 4 * A(n-1)

n                 A(n)

1                 - 6

2                  4 * (-6) = - 24

3                  4 * (-24) = - 96

4                  4 * (-96) = - 384

5                  4 * (-384) = - 1536

6                  4 * ( -1536) = -6144

So, the first six terms are: -6, - 24, - 96, - 384, - 1536, - 6144.

2) Question 2: Find an equation for the nth term of the arithmetic sequence.
-15, -6, 3, 12, ...

option D) - 15 + 9(n - 1)

Explanation:

1. find the difference between the consecutive terms:

-6 - (-15) = -6 + 15 = 9
3 - (-6) = 3 + 6 = 9
12 - 3 = 9

So, the difference is 9, and you can find any term adding 9 to the previous.

2. Since the first term is - 15, you have:

First term, A1 = - 15 + 9(0) = - 15
Second term, A2 =  - 15 + 9(1) = - 6
Third term, A3 = -15 + 9(2) = - 15 + 18 = 3
Fourth term, A4 = - 15 + 9(3) = - 15 + 27 = 12

3. So, the general formula is An = - 15 + 9 (n - 1), which is the option D)

3) Question 3. Find an equation for the nth term of the arithmetic sequence A14 = - 33, A15 = 9.

option B) An = - 579 + 42(n - 1)

Explanation:

1) Find the difference: 9 - (-33) = 9 + 33 = 42

2) A15 = A1 + 42 * (15 - 1)

=> A1 = A15 - 42(15 - 1)

A1 = A15 - 42(14)

A1 = 9 - 588 = - 579

Therefore, the formula es An = - 579 + 42(n - 1)

4) Question 4. Determine whether the sequence converges or diverges. If it converges, give the limit.

48, 8, 4/3, 2/9, ...

the sequence converges to 288/5

Explanation:

That is a geometric sequence.

The ratio is 1/6: 8/48 = 1/6; (4/3) / 8 = 4/24 = 1/6; (2/9)/(4/3) = 6/36 = 1/6.

The convergence criterium is that if |ratio| < 1 then the series, this is the sum of all the terms, converge to: A1 / (1 - ratio)

Then the limit 48 / (1 - 1/6) = 48 / (5/6) = 48*6 / 5 = 288/5

5) Question 5. Find an equation for the nth term of the sequence.

-3, -12, - 48, -192

- 3 * (4)^(n-1)

Explanation: clearly any term (from the second) is the previous term multiplied by 4.

The first term is  -3
The second term is -3(4) = - 12
The third term is -3(4)(4)= - 48
The fourth term is - 3 (4)(4)(4) = - 192

So, the general formula for the nth term is -3 * 4^ (n-1)

6) Question 6. Find an equation for the nth term of a geometric sequence where the second and fifth terms are -21 and 567, respectively.

An =7 * (-3)^(n-1)

Explanation:

1) The fith term is the second term * (ratio)^3: A5 = A3 * (r)^3

2)  A5 = 567, A2 = - 21 => r^3 = A5 / A2 = - 567 / 21 = - 27

=> r = ∛(-27) = - 3

3) So the first term is A1 = A2 / r = -21 / -3 = 7

4) The general formula is

An =7 * (-3)^(n-1)

7) Question 7. Write the sum using summation notation, assuming the suggested pattern continues.

5 - 15 + 45 - 135 + ...

option B) summation of five times negative three to the power of n from n equals zero to infinity

Explanation:

5 = 5
-15 = 5 (-3)
45 = 5(-3)^2
-135 = 5(-3)^3

=> 5 + 5(-3) + 5(-3)^2 + 5(-3)^3+

Using the summation notation that is:

∞
∑ (5)(-3)^n
n=0

Which means summation of five times negative three to the power of n from n equals zero to infinity

8) Question 8. Write the sum using summation notation, assuming the suggested pattern continues.
-9 - 3 + 3 + 9 + ... + 81

option A) summation of the quantity negative nine plus six n from n equals zero to fifteen

Explanation:

Find the difference:

-3 - (-9) = - 3 + 9 = 6
3 - (-3) = 3 + 3 = 6
9 - 3 = 6

First term: - 9
Second term: - 9 + 6(1)
Third term: - 9 + 6(2)

nth term = - 9 + (n -1)

Summation = [- 9] + [- 9 + 6(1) + [-9 + 6(2)] + [-9 + 6(3) ]+ [-9 + 6(15) ]

Using summation notation:

15
∑ [-9 + 6n]
n=0

which means summation of the quantity negative nine plus six n from n equals zero to fifteen.

9) Question 9. Write the sum using summation notation, assuming the suggested pattern continues.


64 + 81 + 100 + 121 + ... + n2 + ...

A) summation of n squared from n equals eight to infinity

Explanation:

64 = 8^2

81 = 9^2

100 = 10^2

121 = 11^2

n^2

=>

∞

∑ n^2

n=8

which means summation of n squared from n equals eight to infinity

10) Question 10. Find the sum of the arithmetic sequence.
17, 19, 21, 23, ..., 35

260

Explanation:

The difference is 2:

The sum is: 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35.

You can use the formula for the sum of an arithmetic sequence:

(A1 + An) * n / 2 = (17 + 35)*10/2 = 260

11) Question 11. Find the sum of the geometric sequence. 

1, 1/2, 1/4, 1/8, 1/16

option D) 31/16

Explanation:

You can either sum the 5 terms or use the formula for the partial sum of a geometric sequence.

The formula is: Sum = A * ( 1 - r^n) / (1 - r)

Here A = 1, r = 1/2, and n = 5 => Sum = 1 * (1 - (1/2)^5 ) / (1 - 1/2) =

= [ 1 - 1/32] / [1/2] = [31/32] / [1/2] = 31 / 16

1.) 10, 20, 40, . . . = 10, 10(2), 10(4), . . . = 10, 10(2)^1, 10(2)^2, . . . = 10(2)^(n - 1)

2.) -4, 8, -16, . . . = -4, -4(-2), -4(4) = -4, -4(-2)^1, -4(-2)^2, . . . = -4(-2)^(n - 1)

3.) a, a + 4, a + 8, . . . = a, a + 1(4), a + 2(4) = a + 4(n - 1)