MAT pure1 sequences and series

Question 1

What is a sequence?

Answer 1

A sequence is a set of numbers in a specific order. These numbers may form an algebraic pattern.

Question 2

What are the common types of sequences?

Answer 2

1. Increasing sequence: Each term is greater than the previous one.
2. Decreasing sequence: Each term is less than the previous one.
3. Arithmetic sequence: The difference between consecutive terms is constant.
   Example: 2, 5, 8, 11, 14…
4. Geometric sequence: The ratio between consecutive terms is constant.
   Example: 1, 3, 9, 27…
5. Periodic sequence: The sequence repeats at regular intervals.
   Example: 1, 3, -4, 1, 3, -4…

Question 3

What is a series?

Answer 3

A series is the sum of the terms of a sequence, often represented using summation (Σ) notation.

Question 4

Example: What does Σ(k=1 to 10) a_k represent?

Answer 4

It represents the series a_1 + a_2 + a_3 + … + a_10.

Question 5

How are sequences defined deductively?

Answer 5

A deductive definition gives a direct formula for the k-th term of the sequence in terms of k.
Example: a_k = 2k - 3.

Question 6

Example: For the sequence a_k = 2k - 3:
a) Write the first 5 terms.
b) Find the 20th term.

Answer 6

a) First 5 terms: -2, 1, 6, 13, 22.
b) 20th term: a_20 = 2(20) - 3 = 37.

Question 7

How are sequences defined inductively?

Answer 7

An inductive definition specifies how to find a term based on the previous term and includes the first term.
Example: a_(k+1) = 2a_k + 1, with a_1 = 0.

Question 8

Example: For the sequence a_(k+1) = 2a_k + 1 and a_1 = 0:
a) Write the first 6 terms.
b) Find Σ(k=1 to 6) a_k.

Answer 8

a) First 6 terms: 0, 1, 3, 7, 15, 31.
b) Sum: 0 + 1 + 3 + 7 + 15 + 31 = 57.

Question 9

What is an arithmetic sequence?

Answer 9

A sequence where the difference between consecutive terms is constant.
Example: 2, 5, 8, 11…

Question 10

What is a geometric sequence?

Answer 10

A sequence where the ratio between consecutive terms is constant.
Example: 1, 3, 9, 27…

Question 11

What is an arithmetic progression (A.P.)?

Answer 11

An A.P. is a sequence where the difference between consecutive terms is constant, called the common difference (d).

Question 12

What is the formula for the k-th term of an arithmetic progression?

Answer 12

a_k = a + (k - 1)d
Where:
- a is the first term,
- d is the common difference,
- k is the term number.

Question 13

What is the formula for the sum of the first n terms of an arithmetic progression?

Answer 13

S_n = (n / 2) * [2a + (n - 1)d]
Where:
- a is the first term,
- d is the common difference,
- n is the number of terms.

Question 14

How can the sum of the first n terms in an arithmetic progression also be written if the last term l is known?

Answer 14

S_n = (n / 2) * (a + l)
Where:
- a is the first term,
- l is the last term.

Question 15

Example: An arithmetic progression starts 2, 5, 8… with 16 terms. Find the last term and the sum.

Answer 15

1. Last term: a = 2, d = 3, n = 16.
   a_16 = 2 + (16 - 1) * 3 = 47.
2. Sum: S_16 = (16 / 2) * [2 * 2 + (16 - 1) * 3]
   S_16 = 8 * 49 = 392.

Question 16

Example: An arithmetic series has first term 3 and the sum of the first 20 terms is 288. Find the common difference.

Answer 16

Use S_n = (n / 2) * [2a + (n - 1)d].
a = 3, S_20 = 288

288 = (20 / 2) * [2 * 3 + (20 - 1)d]
288 = 10 * (6 + 19d)
288 = 60 + 190d
d = 1.2

Question 17

Example: An arithmetic series has common difference -0.5 and the sum of the first 25 terms is 350. Find the first term.

Answer 17

Use S_n = (n / 2) * [2a + (n - 1)d].
d = -0.5, S_25 = 350

350 = (25 / 2) * [2a + 24(-0.5)]
350 = 25 * (a - 6)
a = 20

Question 18

How do you find the number of terms in an AP if the last term is known?

Answer 18

Rearrange the k-th term formula:
l = a + (n - 1)d
Solve for n:
n = [(l - a) / d] + 1

Question 19

Example: The 5th term of an arithmetic progression is 24, and the 9th term is 4. Find a and d.

Answer 19

Use a_k = a + (k - 1)d.
24 = a + 4d
4 = a + 8d
Subtract equations: -20 = -4d → d = -5
Substituting: a = 44

Question 20

Example: The sum of the terms of an arithmetic sequence with first term 5 and common difference 6 is 616. How many terms are there in the sequence?

Answer 20

Use S_n = (n / 2) * [2a + (n - 1)d].
a = 5, d = 6, S_n = 616

616 = (n / 2) * [10 + 6(n - 1)]
616 = (n / 2) * (6n + 4)
3n^2 + 2n - 616 = 0
Solve: (3n + 44)(n - 14) = 0
Since n must be positive, n = 14
The sequence has 14 terms.

Question 21

What is a geometric progression (GP)?

Answer 21

A sequence where the ratio between consecutive terms is constant, called the common ratio (r).

Question 22

What is the formula for the k-th term of a geometric progression?

Answer 22

a_k = a * r^(k - 1)
Where:
- a is the first term,
- r is the common ratio,
- k is the term number.

Question 23

What is the formula for the sum of the first n terms of a geometric progression?

Answer 23

S_n = a * (1 - r^n) / (1 - r) for r ≠ 1.

Question 24

What is the formula for the sum to infinity of a geometric progression?

Answer 24

S_infinity = a / (1 - r)
This applies only if -1 < r < 1

Question 25

Example: Find the 6th term, sum of the first 10 terms, and sum to infinity of a geometric progression with first term = 4, common ratio = 1/2, to 3 d.p. if needed.

Answer 25

1. 6th term: a_6 = 4 * (1/2)^(6 - 1) = 4 / 32 = 1/8
2. S_10 = 4 * (1 - (1/2)^10) / (1 - 1/2) ≈ 7.992
3. S_infinity = 4 / (1 - 1/2) = 8

Question 26

What happens to a geometric progression if |r| > 1?

Answer 26

The terms grow larger or alternate between large positive and negative values, and the series diverges.

Question 27

Example: The sum of a geometric series with first term 3 and common ratio −2 is 513. Find the number of terms in the series.

Answer 27

r = -2, S_n = 513, find n.

1. Use S_n = a * (1 - r^n) / (1 - r).
2. Solve: 3 * (1 - (-2)^n) / (1 - (-2)) = 513
   1 - (-2)^n = 512
   (-2)^n = -512
3. n = 9

Question 28

How do you find the sum of a geometric progression with negative common ratio?

Answer 28

Use the same formula S_n = a * (1 - r^n) / (1 - r), but ensure correct handling of negative powers and signs.

Question 29

How can you find the number of terms in a geometric progression?

Answer 29

If S_n or a specific term is given:
n = log(S_n / a) / log(r)
Ensure you handle |r| and use absolute values if necessary.

Question 30

Example: Carlos saves £100 in the first year and increases by 10% each year. When does his total savings exceed £1000?

Answer 30

1. Use S_n = 100 * (1.1^n - 1) / (1.1 - 1).
2. Solve: 100 * (1.1^n - 1) > 1000.
   1.1^n > 2.
   log(1.1^n) > log(2).
   n > log(2) / log(1.1) ≈ 7.27.
3. Carlos’s savings exceed £1000 after 8 years.

Question 31

What is a binomial expansion?

Answer 31

The binomial expansion expresses (a + b)^n as a series of terms using powers of a and b, where n is a positive integer.

Question 32

What is Pascal’s triangle, and how is it used in binomial expansions?

Answer 32

Pascal’s triangle is a triangular array of numbers where:
- Each entry is the sum of the two numbers above it.
- The n-th row gives the coefficients for (a + b)^n.

Question 33

Example: Expand (p + q)^5 using Pascal’s triangle.

Answer 33

Coefficients from Pascal’s triangle: 1, 5, 10, 10, 5, 1.
Expansion:
(p + q)^5 = p^5 + 5p^4q + 10p^3q^2 + 10p^2q^3 + 5pq^4 + q^5.

Question 34

What is the general formula for binomial coefficients?

Answer 34

C(n, r) = n! / [r! * (n - r)!]
Where:
- n is the total number of terms,
- r is the term number.

Question 35

How do you expand (a + b)^n using the binomial theorem?

Answer 35

(a + b)^n = Σ [C(n, r) * a^(n-r) * b^r], where r ranges from 0 to n.

Question 36

Example: Expand (x + 2)^6

Answer 36

1. Coefficients from Pascal’s triangle: 1, 6, 15, 20, 15, 6, 1.
2. Expansion:
   (x + 2)^6 = x^6 + 6x^5(2) + 15x^4(2^2) + 20x^3(2^3) + 15x^2(2^4) + 6x(2^5) + 2^6.
   = x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64.

Question 37

Example: Find the coefficient of x^5 in the expansion of (3x - 2)^8

Answer 37

1. The term is given by T(r+1) = C(8, r) * (3x)^(8-r) * (-2)^r
2. For x^5, 8 - r = 5 → r = 3
3. T(4) = C(8, 3) * (3x)^5 * (-2)^3
   = 56 * 243x^5 * (-8)
   Coefficient = -108864

Question 38

What is the formula for the binomial expansion of (1 + x)^n for small x?

Answer 38

The first few terms of a binomial expansion make a good approximation when when |x| &laquo_space;1:

(1 + x)^n ≈ 1 + nx + [n(n-1)x^2]/2! + [n(n-1)(n-2)x^3]/3! + …

Question 39

Example: Approximate (1.02)^30 to 4 decimal places.

Answer 39

1. Use (1 + x)^n ≈ 1 + nx + [n(n-1)x^2]/2! + [n(n-1)(n-2)x^3]/3!
2. x = 0.02, n = 30.
   (1.02)^30 ≈ 1 + 30(0.02) + [30(29)(0.02)^2]/2 + [30(29)(28)(0.02)^3]/6
   ≈ 1 + 0.6 + 8.7 × 10^-3 + 1.68 × 10^-4
   ≈ 1.6087