Sequences and iteration

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	5	11	19	29	41
   1st difference		6	8	10	12
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   4, 7, 10, 13, 16
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 3, and b + c = 4, so c = 1
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² + 3n + 1

Check. Check the 3rd term: 3² + 33 + 1. The 3rd term of the sequence is 19.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	3	10	21	36	55
   1st difference		7	11	15	19
   2nd difference			4	4	4

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 4, so a = 2
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   1, 2, 3, 4, 5
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 1, and b + c = 1, so c = 0
5. Write the nth term. Put a, b and c into an n² + bn + c.
   2n² + n

Check. Check the 3rd term: 23² + 3. The 3rd term of the sequence is 21.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	5	12	25	44	69
   1st difference		7	13	19	25
   2nd difference			6	6	6

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 6, so a = 3
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   2, 0, −2, −4, −6
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = -2, and b + c = 2, so c = 4
5. Write the nth term. Put a, b and c into an n² + bn + c.
   3n² − 2n + 4

Check. Check the 3rd term: 33² &mi3us; 23 + 4. The 3rd term of the sequence is 25.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	0	3	8	15	24
   1st difference		3	5	7	9
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   −1, −1, −1, −1, −1
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 0, and b + c = -1, so c = -1
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² − 1

Check. Check the 3rd term: 3² &mi3us; 1. The 3rd term of the sequence is 8.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	4	6	6	4	0
   1st difference		2	0	−2	−4
   2nd difference			−2	−2	−2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = -2, so a = -1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   5, 10, 15, 20, 25
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 5, and b + c = 5, so c = 0
5. Write the nth term. Put a, b and c into an n² + bn + c.
   −n² + 5n

Check. Check the 3rd term: &mi3us;3² + 53. The 3rd term of the sequence is 6.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	5	11	21	35	53
   1st difference		6	10	14	18
   2nd difference			4	4	4

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 4, so a = 2
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   3, 3, 3, 3, 3
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 0, and b + c = 3, so c = 3
5. Write the nth term. Put a, b and c into an n² + bn + c.
   2n² + 3

Check. Check the 3rd term: 23² + 3. The 3rd term of the sequence is 21.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	3	2	3	6	11
   1st difference		−1	1	3	5
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   2, −2, −6, −10, −14
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = -4, and b + c = 2, so c = 6
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² − 4n + 6

Check. Check the 3rd term: 3² &mi3us; 43 + 6. The 3rd term of the sequence is 3.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	5	9	15	23	33
   1st difference		4	6	8	10
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   4, 5, 6, 7, 8
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 1, and b + c = 4, so c = 3
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² + n + 3

Check. Check the 3rd term: 3² + 3 + 3. The 3rd term of the sequence is 15.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	5	9	15	23	33
   1st difference		4	6	8	10
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   4, 5, 6, 7, 8
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 1, and b + c = 4, so c = 3
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² + n + 3

Check. Check the 3rd term: 3² + 3 + 3. The 3rd term of the sequence is 15.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Substitute n = 10 into the nth term.

1. Write the nth term. The nth term is n² + 3n + 1.
   n² + 3n + 1
2. Substitute the position. Replace n with 10.
   1 × 10² + 3 × 10 + 1
3. Work out the value. Use the order of operations.
   = 131

Check. Check the sequence by counting on: the term number 10 is 131.

Watch out. Put the position number, not the term, in place of n.

Method. Substitute n = 12 into the nth term.

1. Write the nth term. The nth term is 2n² − n.
   2n² − n
2. Substitute the position. Replace n with 12.
   2 × 12² − 1 × 12 + 0
3. Work out the value. Use the order of operations.
   = 276

Check. Check the sequence by counting on: the term number 12 is 276.

Watch out. Put the position number, not the term, in place of n.

Method. Substitute n = 5 into the nth term.

1. Write the nth term. The nth term is n² + 2n.
   n² + 2n
2. Substitute the position. Replace n with 5.
   1 × 5² + 2 × 5 + 0
3. Work out the value. Use the order of operations.
   = 35

Check. Check the sequence by counting on: the term number 5 is 35.

Watch out. Put the position number, not the term, in place of n.

Method. Find the first differences, then the differences of those.

1. First differences. Subtract each term from the next.
   5, 7, 9, 11
2. Second differences. Subtract each first difference from the next.
   2, 2, 2
3. Write the answer. The second difference is constant.
   2

Check. The second difference 2 = 2a, so a = 1.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 3. Each term is 2 times the one before.
   a = 3, r = 2
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 3 × 2^{(n − 1)}
3. Substitute n. Put n = 5, so the power is 4.
   T₅ = 3 × 2⁴
4. Work out the term. Multiply the first term by the ratio that many times.
   48

Check. Multiply the terms one at a time to confirm the 5th term is 48.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 2. Each term is 3 times the one before.
   a = 2, r = 3
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 2 × 3^{(n − 1)}
3. Substitute n. Put n = 4, so the power is 3.
   T₄ = 2 × 3³
4. Work out the term. Multiply the first term by the ratio that many times.
   54

Check. Multiply the terms one at a time to confirm the 4th term is 54.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 5. Each term is -2 times the one before.
   a = 5, r = -2
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 5 × -2^{(n − 1)}
3. Substitute n. Put n = 4, so the power is 3.
   T₄ = 5 × -2³
4. Work out the term. Multiply the first term by the ratio that many times.
   −40

Check. Multiply the terms one at a time to confirm the 4th term is −40.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 1. Each term is 10 times the one before.
   a = 1, r = 10
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 1 × 10^{(n − 1)}
3. Substitute n. Put n = 3, so the power is 2.
   T₃ = 1 × 10²
4. Work out the term. Multiply the first term by the ratio that many times.
   100

Check. Multiply the terms one at a time to confirm the 3th term is 100.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 4. Each term is 2 times the one before.
   a = 4, r = 2
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 4 × 2^{(n − 1)}
3. Substitute n. Put n = 6, so the power is 5.
   T₆ = 4 × 2⁵
4. Work out the term. Multiply the first term by the ratio that many times.
   128

Check. Multiply the terms one at a time to confirm the 6th term is 128.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 6. Each term is 1/2 times the one before.
   a = 6, r = 1/2
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 6 × 1/2^{(n − 1)}
3. Substitute n. Put n = 4, so the power is 3.
   T₄ = 6 × 1/2³
4. Work out the term. Multiply the first term by the ratio that many times.
   3/4

Check. Multiply the terms one at a time to confirm the 4th term is 3/4.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 81. Each term is 1/3 times the one before.
   a = 81, r = 1/3
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 81 × 1/3^{(n − 1)}
3. Substitute n. Put n = 3, so the power is 2.
   T₃ = 81 × 1/3²
4. Work out the term. Multiply the first term by the ratio that many times.
   9

Check. Multiply the terms one at a time to confirm the 3th term is 9.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 7. Each term is 1 times the one before.
   a = 7, r = 1
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 7 × 1^{(n − 1)}
3. Substitute n. Put n = 9, so the power is 8.
   T₉ = 7 × 1⁸
4. Work out the term. Multiply the first term by the ratio that many times.
   7

Check. Multiply the terms one at a time to confirm the 9th term is 7.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Divide each term by the term before it.

1. Divide. Divide the second term by the first term.
   12 ÷ 3 = 4
2. Check. Check with the next pair: 48 ÷ 12 = 4.
   48 ÷ 12 = 4
3. Write the answer.
   4

Check. Every term is 4 times the one before.

Method. The first term is 5 and the ratio is 2.

1. Identify a and r. a = 5 and r = 2.
   a = 5, r = 2
2. Write the rule. Use the formula ar⁽ⁿ⁻¹⁾.
   T_n = 5 × 2^{(n − 1)}
3. Write the answer.
   5 × 2^{(n − 1)}

Check. The 3rd term is 5 × 2² = 20.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 100. Each term is 3/5 times the one before.
   a = 100, r = 3/5
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 100 × 3/5^{(n − 1)}
3. Substitute n. Put n = 4, so the power is 3.
   T₄ = 100 × 3/5³
4. Work out the term. Multiply the first term by the ratio that many times.
   108/5

Check. Multiply the terms one at a time to confirm the 4th term is 108/5.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Write 96 as a multiple of the first term.

1. Divide by the first term. 96 ÷ 3 = 32.
   96 ÷ 3 = 32
2. Write 32 as a power of 2. 32 = 2⁵, so 96 = 3 × 2⁵.
   3 × 2⁵
3. Write the answer. The power 5 means it is term number 6.
   Yes, 96 is the 6th term

Check. Term 6 = 3 × 2⁵ = 96.

Method. Divide the 4th term by the 3rd to find r, then work back to the first term.

1. Find the ratio. r = 40 ÷ 20 = 2.
   r = 2
2. Work back to the first term. The 3rd term is a × 2² = 4a = 20, so a = 5.
   4a = 20, so a = 5
3. Write the answer.
   r = 2, a = 5

Check. 5, 10, 20, 40: the 3rd term is 20 and the 4th is 40.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 3. Each term is 2 times the one before.
   a = 3, r = 2
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 3 × 2^{(n − 1)}
3. Substitute n. Put n = 10, so the power is 9.
   T₁₀ = 3 × 2⁹
4. Work out the term. Multiply the first term by the ratio that many times.
   1536

Check. Multiply the terms one at a time to confirm the 10th term is 1536.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Square 10.

1. Write the rule. The nth square number is n × n, so n = 10.
   n = 10
2. Square. Ten multiplied by itself.
   10 × 10 = 100

Check. The square numbers are 1, 4, 9, …, 100.

Method. Cube 6.

1. Cube. 6 × 6 × 6 = 216
   216
2. Write the answer.
   216

Check. The cube numbers are 1, 8, 27, 64, 125, 216.

Method. Use n(n + 1)/2 with n = 8.

1. Substitute. 8 × 9 ÷ 2
   72 ÷ 2
2. Work out. 72 ÷ 2 = 36
   36

Check. Add the numbers 1 to 8: 36.

Method. Use n(n + 1)/2 with n = 20.

1. Substitute. 20 × 21 ÷ 2
   420 ÷ 2
2. Work out. 420 ÷ 2 = 210
   210

Check. Triangular numbers grow faster as n gets bigger.

Method. Add the two terms before to get each new term.

1. Write the terms. The first terms are 1, 1, 2, 3, 5.
   1, 1, 2, 3, 5, 8, 13, 21
2. Count to the 8th term. The 6th term is 8, the 7th is 13, the 8th is 21.
   8 + 13 = 21
3. Write the answer.
   21

Check. 21 = 8 + 13.

Method. Add the two previous terms.

1. Add. 5 + 8 = 13
   13
2. Write the answer.
   13

Check. Then 8 + 13 = 21.

Method. Each shape adds one more row of dots than the one before.

1. Find the differences. The increases are 2, 3, 4, so the next increase is 5.
   3 + 2 = 5, 6 + 4 = 10 is the pattern
2. Add the next row. 10 + 5 = 15.
   10 + 5 = 15

Check. The 6th triangular number is 21.

Method. Look at the differences and the terms.

1. Find the differences. The differences go up by 2 each time.
   4, 6, 8, 10, 12
2. Find the next term. The next term is 30 + 12.
   30 + 12 = 42
3. Explain. The terms are 1 × 2, 2 × 3, 3 × 4, … so they are n(n + 1).
   n(n + 1)

Check. The nth term is n(n + 1), which is twice a triangular number.

Method. Each term is double the one before, which is geometric.

1. Find the ratio. 4 ÷ 2 = 2, and 8 ÷ 4 = 2.
   r = 2
2. Find the 5th term. Double 16.
   16 × 2 = 32

Check. The sequence is geometric with ratio 2.

Method. Test each number.

1. Test 64. 8 × 8 = 64, and 4 × 4 × 4 = 64.
   64 = 8² = 4³
2. Write the answer. 27 is a cube but not a square; 36 and 100 are squares only.
   64

Check. The only number that is both is 64.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 1.0.
   x_n+1 = 1 + 1/x_n
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 1.0000
   x₁ = 2.0000
   x₂ = 1.5000
   x₃ = 1.6667
   …
3. Read off the answer. The values settle down to 1.62 when rounded to 2 d.p.
   x₆ ≈ 1.6154
4. Write the answer. Round x₆ to 2 decimal places.
   1.62

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 1.0.
   x_n+1 = 1 + 2/x_n
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 1.0000
   x₁ = 3.0000
   x₂ = 1.6667
   x₃ = 2.2000
   …
3. Read off the answer. The values settle down to 2.0 when rounded to 1 d.p.
   x₆ ≈ 1.9767
4. Write the answer. Round x₆ to 1 decimal places.
   2.0

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 1.0.
   x_n+1 = (x_n + 2/x_n)/2
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 1.0000
   x₁ = 1.5000
   x₂ = 1.4167
   x₃ = 1.4142
   …
3. Read off the answer. The values settle down to 1.414 when rounded to 3 d.p.
   x₄ ≈ 1.4142
4. Write the answer. Round x₄ to 3 decimal places.
   1.414

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 1.0.
   x_n+1 = (x_n + 1)^(1/3)
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 1.0000
   x₁ = 1.2599
   x₂ = 1.3123
   x₃ = 1.3224
   …
3. Read off the answer. The values settle down to 1.32 when rounded to 2 d.p.
   x₈ ≈ 1.3247
4. Write the answer. Round x₈ to 2 decimal places.
   1.32

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. An iteration converges only if the values settle down to one number.

1. Write the rule. Start with x₀ = 0.0.
   x_n+1 = 3 − 2x_n
2. Work out several values. Put each value into the rule.
   0.00, 3.00, −3.00, 9.00
3. Look at the pattern. The values get bigger in size and alternate in sign, so they do not settle down.
   diverges
4. Write the answer.
   No, it diverges.

Check. Each value is about twice the size of the one before, so the values grow without limit.

Watch out. If the rule multiplies by a number bigger than 1 in size, the values grow.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 2.0.
   x_n+1 = 2/(x_n + 1)
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 2.0000
   x₁ = 0.6667
   x₂ = 1.2000
   x₃ = 0.9091
   …
3. Read off the answer. The values settle down to 1.01 when rounded to 2 d.p.
   x₆ ≈ 1.0118
4. Write the answer. Round x₆ to 2 decimal places.
   1.01

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 2.0.
   x_n+1 = 1 + 1/x_n
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 2.0000
   x₁ = 1.5000
   x₂ = 1.6667
   x₃ = 1.6000
   …
3. Read off the answer. The values settle down to 1.62 when rounded to 2 d.p.
   x₅ ≈ 1.6154
4. Write the answer. Round x₅ to 2 decimal places.
   1.62

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. An iteration converges only if the values settle down to one number.

1. Write the rule. Start with x₀ = 0.5.
   x_n+1 = 3 − 2x_n
2. Work out several values. Put each value into the rule.
   0.50, 2.00, −1.00, 5.00, −7.00
3. Look at the pattern. The values get bigger in size and alternate in sign, so they do not settle down.
   diverges
4. Write the answer.
   No, it diverges.

Check. Each value is about twice the size of the one before, so the values grow without limit.

Watch out. If the rule multiplies by a number bigger than 1 in size, the values grow.

Method. Multiply both sides by x, then collect terms on one side.

1. Multiply by x. This removes the fraction.
   x² = x + 1
2. Move the terms. Subtract x and 1 from both sides.
   x² − x − 1 = 0
3. Write the answer.
   x² − x − 1 = 0

Check. This is the equation the iteration solves, so its solutions are the fixed points.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 10.0.
   x_n+1 = 0.5x_n + 3
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 10.0000
   x₁ = 8.0000
   x₂ = 7.0000
   x₃ = 6.5000
   …
3. Read off the answer. The values settle down to 6.1 when rounded to 1 d.p.
   x₆ ≈ 6.0625
4. Write the answer. Round x₆ to 1 decimal places.
   6.1

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 0.0.
   x_n+1 = 2 + 0.5x_n
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 0.0000
   x₁ = 2.0000
   x₂ = 3.0000
   x₃ = 3.5000
   …
3. Read off the answer. The values settle down to 3.88 when rounded to 2 d.p.
   x₅ ≈ 3.8750
4. Write the answer. Round x₅ to 2 decimal places.
   3.88

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	3	7	13	21	31
   1st difference		4	6	8	10
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   2, 3, 4, 5, 6
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 1, and b + c = 2, so c = 1
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² + n + 1

Check. Check the 3rd term: 3² + 3 + 1. The 3rd term of the sequence is 13.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Substitute n = 8 into the nth term.

1. Write the nth term. The nth term is 2n² + 1.
   2n² + 1
2. Substitute the position. Replace n with 8.
   2 × 8² + 0 × 8 + 1
3. Work out the value. Use the order of operations.
   = 129

Check. Check the sequence by counting on: the term number 8 is 129.

Watch out. Put the position number, not the term, in place of n.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 2. Each term is 3 times the one before.
   a = 2, r = 3
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 2 × 3^{(n − 1)}
3. Substitute n. Put n = 6, so the power is 5.
   T₆ = 2 × 3⁵
4. Work out the term. Multiply the first term by the ratio that many times.
   486

Check. Multiply the terms one at a time to confirm the 6th term is 486.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Divide each term by the one before.

1. Divide. 40 ÷ 80 = 1/2.
   1/2
2. Check. 20 ÷ 40 = 1/2.
   1/2
3. Write the answer.
   1/2

Check. Each term is half the one before.

Method. Use n(n + 1)/2 with n = 7.

1. Substitute. 7 × 8 ÷ 2 = 56 ÷ 2.
   56 ÷ 2
2. Work out. 56 ÷ 2 = 28
   28

Check. 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.

Method. Add the two terms before.

1. Add. 11 + 18 = 29.
   29
2. Write the answer.
   29

Check. 18 = 7 + 11, so the rule is correct.

Method. Find each number and add.

1. Square number. 5² = 25.
   25
2. Cube number. 4³ = 64.
   64
3. Add. 25 + 64 = 89
   89

Check. The square of 5 plus the cube of 4 is 89.

Method. Use the calculator to repeat the rule. Keep the full value each time, and round only at the end.

1. Write the iteration rule. Each new value is found from the rule and the one before it. Start with x₀ = 1.0.
   x_n+1 = 1 + 2/x_n
2. Work out the values. Put each value into the rule to get the next one. Keep more decimal places than you need.
   x₀ = 1.0000
   x₁ = 3.0000
   x₂ = 1.6667
   x₃ = 2.2000
   …
3. Read off the answer. The values settle down to 2.0 when rounded to 1 d.p.
   x₅ ≈ 2.0476
4. Write the answer. Round x₅ to 1 decimal places.
   2.0

Check. When the values stop changing in the first few decimal places, you have the answer.

Watch out. Do not round each value as you go, or the answer may be wrong. Use the full calculator display.

Method. Find the differences.

1. First differences. The differences are 2, 4, 8.
   2, 4, 8
2. Second differences. The second differences are 2, 4, which are not equal.
   2, 4
3. Write the answer. A quadratic sequence has equal second differences.
   Its second differences are not equal

Check. Equal second differences are the test for a quadratic sequence.

Method. Substitute n = 4 into the nth term.

1. Write the nth term. The nth term is n² − 3n.
   n² − 3n
2. Substitute the position. Replace n with 4.
   1 × 4² − 3 × 4 + 0
3. Work out the value. Use the order of operations.
   = 4

Check. Check the sequence by counting on: the term number 4 is 4.

Watch out. Put the position number, not the term, in place of n.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4	5
   term	6	10	16	24	34
   1st difference		4	6	8	10
   2nd difference			2	2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   5, 6, 7, 8, 9
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 1, and b + c = 5, so c = 4
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² + n + 4

Check. Check the 3rd term: 3² + 3 + 4. The 3rd term of the sequence is 16.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Substitute n = 10 into the nth term.

1. Write the nth term. The nth term is n² − 4n + 5.
   n² − 4n + 5
2. Substitute the position. Replace n with 10.
   1 × 10² − 4 × 10 + 5
3. Work out the value. Use the order of operations.
   = 65

Check. Check the sequence by counting on: the term number 10 is 65.

Watch out. Put the position number, not the term, in place of n.

Method. Use the geometric rule with the first term and the common ratio.

1. Identify the first term and the common ratio. The first term is 5. Each term is 3 times the one before.
   a = 5, r = 3
2. Write the rule. The nth term is a × r^{(n − 1)}.
   T_n = 5 × 3^{(n − 1)}
3. Substitute n. Put n = 4, so the power is 3.
   T₄ = 5 × 3³
4. Work out the term. Multiply the first term by the ratio that many times.
   135

Check. Multiply the terms one at a time to confirm the 4th term is 135.

Watch out. The power of the ratio is one less than the term number. The 1st term has power 0.

Method. Divide the 3rd term by the 2nd term.

1. Divide. r = 6 ÷ 12.
   6 ÷ 12
2. Simplify. 6 ÷ 12 = 1/2.
   1/2
3. Write the answer.
   1/2

Check. 12 × 1/2 = 6.

Method. Add the two terms before each new term.

1. Write the terms. 3, 4, 7, 11, then 7 + 11 = 18, then 11 + 18 = 29.
   18, 29
2. Write the answer. The 6th term is 29.
   29

Check. The sequence goes 3, 4, 7, 11, 18, 29.

Method. Use n(n + 1)/2 with n = 12.

1. Substitute. 12 × 13 ÷ 2 = 156 ÷ 2.
   156 ÷ 2
2. Work out. 156 ÷ 2 = 78
   78

Check. 1 + 2 + … + 12 = 78.

Method. Work out each number and add them.

1. Square number. 4² = 16.
   16
2. Cube number. 3³ = 27.
   27
3. Add. 16 + 27 = 43
   43

Check. 16 + 27 = 43 as required.

Method. Use the second difference to find the n² part, then the linear part from what is left.

1. Write the terms and the differences. Find the difference between each pair of terms, then the difference of those differences.

   n	1	2	3	4
   term	2	7	14	23
   1st difference		5	7	9
   2nd difference			2	2

2. Find the second difference. The second difference is constant. For a quadratic it equals 2a.
   2a = 2, so a = 1
3. Subtract a n² from each term. This leaves a linear sequence, which has a constant first difference.
   1, 3, 5, 7
4. Find b and c from the remaining sequence. The first difference of the remainder is b, and its first term is b + c.
   b = 2, and b + c = 1, so c = -1
5. Write the nth term. Put a, b and c into an n² + bn + c.
   n² + 2n − 1

Check. Check the 3rd term: 3² + 23 &mi3us; 1. The 3rd term of the sequence is 14.

Watch out. Only the second difference of a quadratic sequence is constant. If the second differences are not equal, the sequence is not quadratic.

Method. Substitute n = 4 into the nth term.

1. Write the nth term. The nth term is 3n² + 2n.
   3n² + 2n
2. Substitute the position. Replace n with 4.
   3 × 4² + 2 × 4 + 0
3. Work out the value. Use the order of operations.
   = 56

Check. Check the sequence by counting on: the term number 4 is 56.

Watch out. Put the position number, not the term, in place of n.

Method. Divide each term by the one before.

1. Divide. 8 ÷ 16 = 1/2, and 4 ÷ 8 = 1/2.
   1/2
2. Write the answer. The same ratio each time makes it geometric.
   1/2

Check. Each term is half the one before.