Recursive rule

The governing concept is recursive evaluation with an alternating correction term. The sign of (−1)ⁿ changes with the parity of n, so it must be checked at every step. Starting from a₁ = 7: for n = 1, a₂ = 2(7) − 1 = 13; for n = 2, a₃ = 2(13) + 2 = 28; for n = 3, a₄ = 2(28) − 3 = 53; and for n = 4, a₅ = 2(53) + 4 = 110. Thus option C is correct. Odd n gives a negative correction, whereas even n gives a positive correction. Ignoring the alternating sign or applying the wrong n produces the distractor values.

Using n=1, 2, 3 and 4 in the recursive rule, the added values are 3, 8, 17 and 32 respectively. Thus, a_2=9, a_3=17, a_4=34, and a_5=34+32=66. Therefore, the correct answer is 66. The value 69 can result from an incorrect addition in the final step. Exam tip: calculate 2^n and n^2 separately at each step before adding them.

This is a recursive sequence because each term is obtained from the preceding term. For n = 1, subtract 3¹ + 1 = 4, giving a₂ = 196. For n = 2, subtract 3² + 2 = 11, giving a₃ = 185. For n = 3, subtract 3³ + 3 = 30, so a₄ = 185 − 30 = 155. Therefore, option A is correct; the other values result from an arithmetic error in one subtraction.

Substitute n=1, 2, 3, and 4 in the recursive rule. The added values are \(1!+1=2\), \(2!+2=4\), \(3!+3=9\), and \(4!+4=28\). Hence, \(a_2=7\), \(a_3=11\), \(a_4=20\), and \(a_5=20+28=48\). Therefore, 48 is correct. A value such as 50 results from an incorrect addition in one of the steps. Exam tip: to find \(a_5\), apply the rule only for \(n=1\) through \(4\).

Substituting n=1, 2, 3, and 4 into the recursive rule gives the subtracted values 3, 6, 12, and 32. Thus, a_2=90-3=87, a_3=87-6=81, a_4=81-12=69, and a_5=69-32=37. Therefore, 37 is correct. The value 34 can result from an error in the final subtraction; for n=4, 4!=24, not 6. Exam tip: to find a_5, apply the rule only from n=1 through n=4.

Using n=1 in the recursive rule, a_2=4a_1-(1+1)=4×3-2=10. Then, using n=2, a_3=4a_2-(2+1)=4×10-3=37. Hence, 37 is the correct option. The value 39 would result from incorrectly subtracting 1 instead of 3 in the second step. Exam tip: write the current value of n before substituting it at each step.

First, put n=3: a_3=6+2(2)+3^2=19. Then, for n=4, a_4=19+2(6)+4^2=19+12+16=47. Therefore, 47 is correct. Using only the previous term gives an incorrect result because the rule also includes twice a_{n-2} and n^2. Exam tip: substitute the correct value of n at each step and write every term separately.

The given rule uses the two preceding terms, so calculate successively. First, a₃ = 3a₂ − 2a₁ = 3(11) − 2(5) = 23. Next, a₄ = 3(23) − 2(11) = 47. Finally, a₅ = 3(47) − 2(23) = 141 − 46 = 95. Thus option C is correct. The nearby alternatives usually come from using only one earlier term or making an arithmetic error.

Apply the recursive rule successively for n=1, 2, and 3. We get a_2=8+1(1+2)+2^1=13, a_3=13+2(2+2)+2^2=25, and a_4=25+3(3+2)+2^3=48. Hence, 48 is correct. A value such as 45 may result from incorrectly evaluating 2^3 in the final step. Exam tip: to find a_4 from a_1, apply the rule exactly three times.

Apply the recurrence for n = 1, 2, and 3. The first decrement is 1(3) + 2¹ = 5, so a₂ = 180 − 5 = 175. The second is 2(4) + 2² = 12, so a₃ = 163. The third is 3(5) + 2³ = 23, so a₄ = 163 − 23 = 140. Hence option C is correct; the other choices reflect an incorrect value of one decrement.

The recurrence has two parts: double the previous term, then add 3n − 2. To find a₃, use the rule first with n = 1 and then with n = 2. The value of n belongs to the step being performed, while the sequence term on the right is the immediately preceding term.

Starting from a₁ = 9, for n = 1 we get a₂ = 2(9) + 3(1) − 2 = 18 + 1 = 19. For n = 2, use a₂ = 19: a₃ = 2(19) + 3(2) − 2 = 38 + 6 − 2 = 42. Thus the terms needed are 9, 19, and 42, so option C is correct. A common error is to add 3n − 2 to the original term again instead of using the newly calculated a₂.

Use the recursive rule first with \(n=1\) to find \(a_2\): \(a_2=\frac{30}{3}+2(1)^2=10+2=12\). Then use \(n=2\): \(a_3=\frac{12}{3}+2(2)^2=4+8=12\). Therefore, the correct answer is 12. A value such as 14 can result from evaluating the \(2n^2\) term incorrectly. Exam tip: substitute the correct value of \(n\) at every step of a recursive sequence.

First, put n=1: a_2=5+3(5)+1^2=21. Then put n=2: a_3=21+3(5)+2^2=40. Therefore, the correct answer is 40. The term 3a_1 remains 15 at every step; only n^2 changes. Exam tip: In a recursive rule, calculate each preceding term step by step until the required term is reached.

The governing concept is a two-term recurrence: from a₃ onward, each term is formed by adding the two preceding terms and the index-dependent quantity 3n. For n = 3, a₃ = a₂ + a₁ + 3(3) = 4 + 2 + 9 = 15. For n = 4, a₄ = a₃ + a₂ + 3(4) = 15 + 4 + 12 = 31. For n = 5, a₅ = a₄ + a₃ + 3(5) = 31 + 15 + 15 = 61. Therefore option D is correct. The index in 3n must match the term being calculated, and both preceding terms must be included. Omitting one term or using a wrong index explains the other options.

First substitute \(n=3\) in the recursive rule: \(a_3=2(5)+2+3^2=21\). Then, for \(n=4\), \(a_4=2(21)+5+4^2=42+5+16=63\). A value such as 66 can result from an error while calculating \(a_3\). Exam tip: always add the square of the current value of \(n\) separately.

Apply the recursive rule first to a_1=3: a_2=3^2-2(3)+4=9-6+4=7. Then use a_2=7: a_3=7^2-2(7)+4=49-14+4=39. Therefore, the correct answer is 39. A value such as 42 results from an arithmetic error. Exam tip: to find a_3, apply the rule successively to a_1 and then to a_2.

Putting \(n=1\) in the recursive rule gives \(a_2=a_1^2-a_1+2\). Therefore, \(a_2=4^2-4+2=16-4+2=14\). Option 12 can result from an error in evaluating \(4^2\) or in the final addition. Exam tip: to find \(a_2\), substitute the value of \(a_1\) into the rule.

(79-13=66), and (66\div6=11) steps, so the term number is (12). Exam tip: add (1) to the number of steps.

For (a_{12}), (9) is added (11) times, so (a_{12}=107). Exam tip: the rule is applied one less time than the term number.

Use the recursive rule to find the required terms: \(a_2=2(4)+9=17\), \(a_3=2(17)+9=43\), and \(a_4=2(43)+9=95\). Therefore, \(a_4-a_2=95-17=78\), so option D is correct. A common error is to stop at \(a_3\) instead of calculating the required fourth term. Exam tip: in a recursive sequence, generate terms step by step until both requested terms are obtained.