Recursive rule

The recursive rule gives each new term by adding n(n+2) to the preceding term. Thus, C_2=4+2(4)=12, C_3=12+3(5)=27, and C_4=27+4(6)=51. Therefore, the correct answer is 51. A value such as 54 can result from using an incorrect value of n for a step. Exam tip: To find C_4, apply the rule successively for n=2, 3, and 4.

Use the index of the term being calculated in the recursive rule. Thus, \(D_2=3(5)-(2+1)=12\), \(D_3=3(12)-(3+1)=32\), and \(D_4=3(32)-(4+1)=96-5=91\). Therefore, the correct answer is 91. A value such as 95 can result from incorrectly subtracting 1 instead of \(n+1\). Exam tip: write the value of \(n\) at each step before evaluating \(n+1\).

In the recursive rule, each new term is obtained by adding the previous two terms and the current index n. Thus, E_3=2+4+3=9, E_4=4+9+4=17, and E_5=9+17+5=31. Therefore, the correct answer is 31. A value such as 29 may result from omitting +5 in the last step. Exam tip: use the subscript of the term as the value of n at each step.

The recursive rule requires subtracting the value based on the current index at every step. Thus, \(F_2=90-2(3)=84\), \(F_3=84-3(4)=72\), and \(F_4=72-4(5)=52\). Therefore, the correct answer is 52. A value such as 56 may result from subtracting an incorrect value instead of \(4(5)\). Exam tip: to find \(F_4\), substitute \(n=2,3,4\) in order and write each intermediate term.

In a recursive rule, substitute the index of the term currently being found. Thus, \(G_2=2(1)+2(2)+1=7\), \(G_3=2(7)+2(3)+1=21\), and \(G_4=2(21)+2(4)+1=51\). Therefore, the correct answer is \(51\). A value such as 47 can result from adding the \(2n+1\) term incorrectly. Exam tip: at each step, first double the previous term, then substitute the current value of \(n\) and add \(2n+1\).

Option B uses both \(a_{n-1}\) and \(a_{n-2}\), so two starting values, \(a_1\) and \(a_2\), are needed. A and D are first-order rules. Exam tip: check the farthest previous subscript used.

(J_2=37), (J_3=30), (J_4=19), and (J_5=4). In exams, handle both the subtraction and final addition of (1).

(K_3=13), (K_4=34), and (K_5=89). In exams, triple the previous term and subtract the term before it.

The recursive rule forms each new term by adding (5n-4) to the preceding term. Thus, L_2=4+6=10, L_3=10+11=21, L_4=21+16=37, and L_5=37+21=58. Therefore, 58 is correct. The option 55 can result from using an incorrect final increment instead of 21. Exam tip: To find L_5, apply the increments successively for n=2 through n=5.

In a recursive rule, first find the next term from the previous one. \(M_2=4M_1-2=4\times2-2=6\). Then \(M_3=4M_2-3=4\times6-3=21\). Therefore, the correct answer is 21. The value 25 would result from forgetting to subtract 3 in the final step. Exam tip: use the current index \(n\) each time you apply the rule.

In a recursive rule, use the index of the term being calculated at every step. \(R_2=(2+2)\times1-1=3\), \(R_3=(3+2)\times3-1=14\), and \(R_4=(4+2)\times14-1=83\). Therefore, the correct answer is eighty-three. A distractor such as seventy-nine may result from not changing the factor \(n+2\) correctly at each step. Exam tip: calculate \(R_2\) and \(R_3\) in order before finding \(R_4\).

Apply the recursive rule step by step: \(O_2=5+2^2+2(2)=13\), \(O_3=13+2^3+2(3)=27\), and \(O_4=27+2^4+2(4)=51\). Therefore, the correct answer is 51. A value such as 55 can result from an error while evaluating the \(2n\) term. Exam tip: calculate \(2^n\) and \(2n\) separately at each step before adding them.

Apply the recursive rule successively: \(P_2=P_1+3(2)^2-1=11\), \(P_3=11+3(3)^2-1=37\), and \(P_4=37+3(4)^2-1=84\). Hence, the correct answer is \(84\). The option \(81\) may result from incorrectly using \(3n^2-2\) instead of \(3n^2-1\). Exam tip: evaluate \(3n^2-1\) for each value of \(n\) before adding it to the preceding term.

Under the recursive rule, each new term is formed using the previous two terms and the square of its own index. Thus, (Q_3=1+2+3^2=12), (Q_4=12+2+4^2=30), and (Q_5=30+12+5^2=67). Therefore, 67 is correct. A value such as 61 can result from using one of the preceding terms incorrectly. Exam tip: add the square of the current index at every step.

In a recursive rule, each new term is calculated from the preceding term. First, \(U_2=U_1^2-2=3^2-2=7\). Then, \(U_3=U_2^2-3=7^2-3=49-3=46\). Therefore, the correct answer is 46. A value such as 44 could result from subtracting 5 instead of 3 in the final step. Exam tip: write the value of \(n\) separately at every step before subtracting it.

In the recursive rule, use the exponent corresponding to the current term. Thus, \(S_2=75-2^2=71\), \(S_3=71-2^3=63\), and \(S_4=63-2^4=47\). Therefore, 47 is correct. The value 49 results if 2 is incorrectly subtracted instead of \(2^2\) while finding \(S_2\). Exam tip: To reach \(S_4\), substitute \(n=2,3,4\) in order.

Using the recursive rule, first put \(n=2\): \(T_2=(2+2)\times2+2=10\). Next, put \(n=3\): \(T_3=(3+2)\times10+3=53\). Therefore, the correct answer is \(53\). A value such as \(49\) can result from using an incorrect multiplier instead of \(n+2\). Exam tip: substitute the value of \(n\) into the rule first, then use the preceding term.

An arithmetic progression has a constant difference between consecutive terms. Its recursive form is \(a_n=a_{n-1}+d\), so the same fixed number \(d\) is added each time. A geometric progression uses a constant multiplier, not a fixed addition. Exam tip: “constant difference” identifies an AP.

The governing concept is a recursive sequence in which later terms are obtained by adding a quantity depending on n. To connect a₂ with a₅, use the recurrence for n = 3, 4 and 5. The increments are 4(3) − 2 = 10, 4(4) − 2 = 14 and 4(5) − 2 = 18. Therefore a₅ = a₂ + 10 + 14 + 18 = a₂ + 42. Given a₅ = 70, a₂ = 70 − 42 = 28. Thus option D is correct. A forward check gives 28 + 10 = 38, 38 + 14 = 52 and 52 + 18 = 70. The other options result from using the wrong range of n, omitting an increment or making an arithmetic subtraction error.

Using the recursive rule, (b_2=3b_1+2). Therefore, (b_3=3(3b_1+2)+2=9b_1+8). Since (b_3=44), we get (9b_1+8=44), so (9b_1=36) and hence (b_1=4). Substituting 5, 6, or 7 would not give (b_3=44). Exam tip: expand a recursive relation step by step until it connects to the required term.