Recursive rule

At each step, the recursive rule adds \(n^2+2n+1=(n+1)^2\). Thus, \(a_2=2+4=6\), \(a_3=6+9=15\), and \(a_4=15+16=31\). Therefore, 31 is correct. Option 30 may result from incorrectly using 15 instead of 16 for the final increment. Exam tip: quickly recognise \(n^2+2n+1\) as \((n+1)^2\).

The terms are (24,32,\frac{128}{3}), so (a_3=\frac{128}{3}). The correct option should include the fractional value.

Given \(a_1=18\), put \(n=1\) in the recursive rule: \(a_2=a_1+\frac{a_1}{3}=18+\frac{18}{3}=18+6=24\). Hence, 24 is correct. The value 22 would result from incorrectly taking \(18/3\) as 4, whereas it is 6. Exam tip: to find \(a_{n+1}\), substitute the preceding term directly into the rule.

Apply the recursive rule step by step: \(a_3=4\times3-3\times1=9\), \(a_4=4\times9-3\times3=27\), and \(a_5=4\times27-3\times9=81\). Hence, 81 is correct. The close distractor 27 is \(a_4\), not \(a_5\). Exam tip: while using a recurrence, keep track of the two previous terms in the correct order.

This is a second-order recursive sequence because each new term depends on the two preceding terms, not just on one term. The supplied starting values are a₁ = 1 and a₂ = 3. For n = 3, calculate a₃ = 4a₂ − 3a₁ = 4(3) − 3(1) = 12 − 3 = 9. Then substitute a₃ and a₂ when n = 4: a₄ = 4a₃ − 3a₂ = 4(9) − 3(3) = 36 − 9 = 27. Hence option C is correct, and the sequence begins 1, 3, 9, 27. A common mistake is to use a₁ instead of a₂ in the second calculation, reverse the coefficients, or ignore the second previous term. Such errors can produce the other choices, but they do not follow the stated recurrence.

(a_2=1\times3+2=5), (a_3=2\times5+2=12), and (a_4=3\times12+2=38). The correct option should include (38).

The governing idea is successive use of the recurrence with the correct current index at every step. Start with a₁ = 4. For n = 1, the rule gives a₂ = 2a₁ + 1 = 2(4) + 1 = 9. For n = 2, use a₂ to obtain a₃ = 2a₂ + 2 = 2(9) + 2 = 20. For n = 3, calculate a₄ = 2a₃ + 3 = 2(20) + 3 = 43. Therefore option C is correct. The added values are 1, 2, and 3 because those are the indices of the transitions producing a₂, a₃, and a₄. Adding 4 while calculating a₄ applies the rule one step too late. Forgetting the factor 2 or using an incorrect preceding term can explain the other plausible options.

Apply the recursive rule successively for n=1,2,3,4: a_2=50-5=45, a_3=45-7=38, a_4=38-9=29, and a_5=29-11=18. Therefore, 18 is correct. A value such as 20 can result from using an incorrect value of n or subtracting the wrong amount. Exam tip: to reach a_5 from a_1, apply the rule four times.

In an arithmetic progression, the difference \(a_{n+1}-a_n\) is constant. Rule A gives this difference as \(3\), so its common difference is 3. In option C, the difference is \(3n\), which changes with \(n\). Exam tip: a fixed number added at each step is the common difference.

The added values are (3,8,15,24), and after adding them (a_5=50). Exam tip: write values of (n^2+2n) separately.

The governing concept is a recursive subtraction with a changing decrement. To move from a₁ to a₅, apply the rule for n = 1, 2, 3, and 4. The amounts subtracted are n² + n: for n = 1, the amount is 1² + 1 = 2; for n = 2, it is 2² + 2 = 6; for n = 3, it is 3² + 3 = 12; and for n = 4, it is 4² + 4 = 20. Step by step, a₂ = 120 − 2 = 118, a₃ = 118 − 6 = 112, a₄ = 112 − 12 = 100, and a₅ = 100 − 20 = 80. Equivalently, the total deduction is 2 + 6 + 12 + 20 = 40, so 120 − 40 = 80. Thus option B is correct. The decrement is not constant, and subtracting only the last amount or using five transitions would be incorrect.

This is a second-order recurrence: each new term equals the immediately previous term plus twice the term two positions earlier. Start with a₁ = 1 and a₂ = 4. For n = 3, a₃ = a₂ + 2a₁ = 4 + 2(1) = 6. For n = 4, a₄ = a₃ + 2a₂ = 6 + 2(4) = 14. For n = 5, a₅ = a₄ + 2a₃ = 14 + 2(6) = 26. Finally, for n = 6, a₆ = a₅ + 2a₄ = 26 + 2(14) = 54. Therefore option B is correct, and the sequence is 1, 4, 6, 14, 26, 54. The coefficient 2 must multiply the term two positions back, not the newest term. Confusing these positions or using only one previous term can produce the other choices but violates the stated recurrence.

The governing concept is a second-order recursive rule: each new term is calculated from the two immediately preceding terms. Start with a₁ = 5 and a₂ = 8. Then a₃ = 2(8) − 5 = 11, a₄ = 2(11) − 8 = 14, a₅ = 2(14) − 11 = 17, and a₆ = 2(17) − 14 = 20. Finally, a₇ = 2(20) − 17 = 23. Thus option D is correct. The order of the terms matters: the most recent term is doubled, and the term before it is subtracted. The resulting sequence is 5, 8, 11, 14, 17, 20, 23, which also has a constant increase of 3 and independently confirms the result. The other options are earlier terms or arise from an arithmetic or indexing mistake.

The added values are (3,6,11), so (a_4=2+3+6+11=22). Exam tip: calculate both (2^n) and (n) before adding.

The governing concept is repeated use of an index-dependent recurrence. Four transitions are needed to move from a₁ to a₅, so evaluate 2ⁿ − n for n = 1, 2, 3, and 4. The increments are 2¹ − 1 = 1, 2² − 2 = 2, 2³ − 3 = 5, and 2⁴ − 4 = 12. Starting from a₁ = 3 gives a₂ = 3 + 1 = 4, a₃ = 4 + 2 = 6, a₄ = 6 + 5 = 11, and a₅ = 11 + 12 = 23. Equivalently, a₅ = 3 + 1 + 2 + 5 + 12 = 23. Therefore option D is correct. Option C may result from an addition error, while applying the rule five times would incorrectly introduce a transition beyond the requested fifth term.

The governing concept is recursive evaluation: each term is obtained from the preceding term, and the index n must be substituted separately at every transition. Start with a₁ = 6. For n = 1, (-1)¹(2×1) = -2, so a₂ = 6 - 2 = 4. For n = 2, (-1)²(2×2) = 4, so a₃ = 4 + 4 = 8. For n = 3, the change is (-1)³(6) = -6, giving a₄ = 8 - 6 = 2. For n = 4, the change is (+1)(8) = 8, so a₅ = 2 + 8 = 10. Therefore option C is correct. The other choices can result from ignoring the alternating sign, using the wrong index, or stopping before the fourth update.

Substitute n=1, 2, and 3 successively in the recursive rule. a_2=6+(-1)^1(2)=4, a_3=4+(-1)^2(4)=8, and a_4=8+(-1)^3(6)=2. Therefore, the correct answer is 2. Although 4 is a close distractor, it is only the second term. Exam tip: write the value of n at every step, since (-1)^n is negative for odd n and positive for even n.

Apply the recursive rule successively for n=1, 2, and 3: a_2=4+1(2)+2=8, a_3=8+2(3)+2=16, and a_4=16+3(4)+2=30. Therefore, the correct answer is 30. The value 28 results if the final +2 is omitted. Exam tip: to find a_4 from a_1, apply the rule three times.

A recursive rule gives each new term from the term immediately before it. To find the fourth term, begin with the first term and apply the rule for n = 1, then n = 2, and finally n = 3. The subtraction changes at every step because the value of n changes.

For n = 1, subtracting gives 1(2) + 2 = 4, so a₂ = 90 − 4 = 86. For n = 2, subtract 2(3) + 2 = 8, giving a₃ = 86 − 8 = 78. For n = 3, subtract 3(4) + 2 = 14, giving a₄ = 78 − 14 = 64. Equivalently, the total subtraction is 4 + 8 + 14 = 26, so a₄ = 90 − 26 = 64. Therefore option A is correct. The explanation supplied is accurate, although showing the intermediate terms makes the method clearer.

In a recursive rule, each next term is found using the previous term and its index. \(a_2=2(3)+2(1)+1=9\), \(a_3=2(9)+2(2)+1=23\), and \(a_4=2(23)+2(3)+1=53\). Therefore, the correct answer is 53. A value such as 49 can result from using an incorrect value of \(n\) in the term \(2n+1\). Exam tip: write the current value of \(n\) at every step while applying a recursive formula.