Recursive rule

A recursive definition needs both an initial term and a rule connecting each new term to a previous term. Option A provides \(a_1=4\) and \(a_{n+1}=a_n+3\). Option C lacks an initial value. Exam tip: check for both parts.

This is a second-order recursive sequence because each term depends on the two immediately preceding terms. The initial values are a₁ = 2 and a₂ = 5. For n = 3, a₃ = a₂ + 2a₁ = 5 + 2(2) = 9. For n = 4, a₄ = a₃ + 2a₂ = 9 + 2(5) = 19. For n = 5, a₅ = a₄ + 2a₃ = 19 + 2(9) = 37. Therefore option D is correct. The coefficient 2 must multiply the term two positions earlier, not the immediately preceding term. Values 23, 27, and 31 can result from stopping before the fifth term or applying that coefficient to the wrong term. Writing every intermediate term prevents such indexing errors.

In an arithmetic sequence, consecutive terms have a constant difference. Here \(a_n-a_{n-1}=3\), so each new term is obtained by adding 3 to the previous term. Option B multiplies terms and is geometric in nature. Exam tip: for a fixed difference, use “previous term + d”.

Using the recursive rule, each new term is twice the previous term minus the term before it. Thus, \(a_3=2(10)-4=16\), \(a_4=2(16)-10=22\), \(a_5=28\), and \(a_6=34\). Therefore, the correct answer is 34. Although 32 is a close distractor, it is not a term of this sequence. Exam tip: write the next few terms in order when evaluating a recursive sequence.

In a recursive rule, each new term is found by substituting the current value of n into the increment and adding it to the previous term. Thus, a_2=6+(1^2+2×1)=9, a_3=9+(2^2+2×2)=17, and a_4=17+(3^2+2×3)=32. Therefore, 32 is correct. The value 30 may result from incorrectly calculating the third increment instead of using 15. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

Substitute n=1, 2, and 3 in the recursive rule. The quantities subtracted are 2, 6, and 12 respectively. Thus, a_2=90-2=88, a_3=88-6=82, and a_4=82-12=70. Therefore, 70 is correct. The value 72 can result from using an incorrect value for n^2+n in the third step. Exam tip: to find a_4 from a_1, apply the rule three times.

In a recursive rule, each new term is obtained by adding the required power to the previous term. Thus, a_2=1+3^1=4, a_3=4+3^2=13, and a_4=13+3^3=40. Therefore, the correct answer is 40. The value 39 would result from incorrectly using 26 instead of the final increment 27. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

Using n=1, 2, and 3 in the recursive rule, subtract 3, 9, and 27 respectively. Thus, \(a_2=100-3=97\), \(a_3=97-9=88\), and \(a_4=88-27=61\). Therefore, the correct answer is 61. The value 64 would result from subtracting 24 in the last step, but the rule requires subtracting \(3^3=27\). Exam tip: Increase n by 1 for each new term before evaluating the power.

Apply the recursive rule successively for n=1, 2, and 3. We get a_2=8+2(-1)^1+1=7, a_3=7+2(-1)^2+2=11, and a_4=11+2(-1)^3+3=12. Hence, the correct answer is 12. A value such as 10 can result from mishandling the alternating sign in (-1)^n. Exam tip: write the value of n at every step and check the sign separately.

First, put n=1: \(a_2=8+2(-1)^1+1=8-2+1=7\). Then put n=2: \(a_3=7+2(-1)^2+2=7+2+2=11\). Therefore, the correct answer is 11. Choosing 10 would incorrectly treat \((-1)^2\) as negative. Exam tip: in a recursive rule, substitute consecutive values of n carefully while finding each next term.

In the recursive rule, add \(n(n+3)\) to the current term each time. \(a_2=3+1(1+3)=7\), \(a_3=7+2(2+3)=17\), and \(a_4=17+3(3+3)=35\). Hence, the correct answer is 35. The option 36 may result from incorrectly taking the final increase as 19 instead of 18. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).

Given \(a_1=3\) and \(a_{n+1}=a_n+n(n+3)\), first put \(n=1\): \(a_2=3+1(1+3)=7\). Next, put \(n=2\): \(a_3=7+2(2+3)=7+10=17\). Therefore, the correct answer is 17. A result such as 15 can arise from incorrectly evaluating the added term. Exam tip: in a recursive rule, substitute the correct value of \(n\) at every step.

Substitute n=1, 2, and 3 in the recursive rule. The amounts subtracted are 1(1+3)=4, 2(2+3)=10, and 3(3+3)=18. Thus, a_2=120-4=116, a_3=116-10=106, and a_4=106-18=88. Therefore, 88 is correct. An option such as 90 can result from an incorrect subtraction in the final step. Exam tip: to find a_4, apply the rule only for n=1 through n=3.

In a recursive rule, each new term is found from the preceding term. Thus, \(a_2=2(4)+5=13\), \(a_3=2(13)+5=31\), and \(a_4=2(31)+5=67\). Therefore, 67 is correct. A value such as 65 can result from applying the rule incorrectly to the previous term. Exam tip: To find \(a_4\) from \(a_1\), apply the rule three times in order.

The first term is 5, and each next term is obtained by adding 3: \(5+3=8\) and \(8+3=11\). Hence A is correct. In B, terms are multiplied by 3 instead. Exam tip: generate two terms to check a recursive rule.

To find the second term, substitute \(n=1\) and \(a_1=5\) in the recursive rule: \(a_2=3a_1+1+2=3\times5+1+2=18\). Therefore, 18 is correct. Option 16 may result from not adding the \(n+2\) part correctly. Exam tip: while finding \(a_{n+1}\), first substitute the correct value of \(n\).

In an arithmetic progression, each new term is obtained by adding a fixed constant to the previous term. Here, 4 is added every time, so the common difference is constant. Option B multiplies terms by 4, which indicates a geometric progression. Exam tip: look for “+ constant” to identify an AP.

The governing concept is repeated application of a recurrence whose increment changes with n. To reach a₅ from a₁, use n = 1, 2, 3, and 4. First, a₂ = 2 + 1(2)/2 + 1 = 2 + 1 + 1 = 4. Next, a₃ = 4 + 2(3)/2 + 1 = 4 + 3 + 1 = 8. Then, a₄ = 8 + 3(4)/2 + 1 = 8 + 6 + 1 = 15. Finally, a₅ = 15 + 4(5)/2 + 1 = 15 + 10 + 1 = 26. Thus option B is correct. The successive increases are 2, 4, 7, and 11, so this is not a constant-difference progression. Omitting the final +1, treating the triangular expression as fixed, or using the wrong index can lead to the distractors 24, 28, or 30.

A recursive rule must be applied one step at a time. To obtain a₂, use n = 1: a₂ = 2 + 1(2)/2 + 1 = 2 + 1 + 1 = 4. To obtain a₃, use n = 2: a₃ = 4 + 2(3)/2 + 1 = 4 + 3 + 1 = 8. To obtain a₄, use n = 3: a₄ = 8 + 3(4)/2 + 1 = 8 + 6 + 1 = 15. Therefore option B is correct. The successive increments are 2, 4, and 7, so this is not an arithmetic progression with a fixed common difference. Option A could arise from omitting the final 1, while options C and D reflect incorrect substitution or arithmetic. The changing value of n is central to the calculation.

Substituting \(n=1,2,3\) in the recursive rule gives subtractions of \(2,4,7\), respectively. Thus, \(a_2=80-2=78\), \(a_3=78-4=74\), and \(a_4=74-7=67\). Therefore, the correct answer is 67. A value such as 69 can result from incorrectly subtracting 5 instead of 7 in the final step. Exam tip: to find \(a_4\), apply the recurrence only for \(n=1\) through \(n=3\).