Recursive rule

A recursive rule uses the preceding term to obtain the next term. \(R_2=2\times2+2=6\), \(R_3=3\times6+3=21\), and \(R_4=4\times21+4=88\). Hence, 88 is correct. A value such as 82 can result from an error while adding \(4\) in the final step. Exam tip: substitute the current value of \(n\) separately at every step.

Under the recursive rule, each new term is obtained by adding \(2^n+n\) to the previous term. \(O_2=4+2^2+2=10\), \(O_3=10+2^3+3=21\), and \(O_4=21+2^4+4=41\). Therefore, the correct answer is 41. A value such as 45 may result from incorrectly adding an extra 4 while finding the fourth term. Exam tip: write the value of \(n\) separately at every step before evaluating the power.

Substituting n=2,3,4,5 in the recursive rule gives increments of 5, 9, 13, and 17. Thus, P_2=8, P_3=17, P_4=30, and P_5=47. Therefore, 47 is correct. A value such as 45 may result from an incorrect final increment. Exam tip: in a recursive sequence, calculate each term successively from the preceding term.

Each new term is obtained from the previous two terms using the recursive rule. Q_3=5+2(2)+3=12, then Q_4=12+2(5)+4=26, and Q_5=26+2(12)+5=55. Hence, the correct answer is 55. A value such as 57 can result from incorrectly handling 2Q_{n-2} or the added n. Exam tip: substitute n=3, 4, and 5 step by step.

In a recursive rule, first find each next term from the preceding term. R_2=R_1^2+2=1^2+2=3. Then R_3=R_2^2+3=3^2+3=12. Therefore, 12 is correct. A value such as 10 can result from not adding the correct term number after squaring R_2. Exam tip: write the new value of n at every step.

The recursive rule gives each term from the preceding term. Thus, \(S_2=100-2^2=96\), \(S_3=96-3^2=87\), and \(S_4=87-4^2=87-16=71\). Therefore, 71 is correct. A value such as 69 can result from using incorrect squares in the steps. Exam tip: since the sequence starts at \(S_1\), subtract \(2^2,3^2\), and \(4^2\) to find \(S_4\), not \(1^2\).

In the recursive rule, substitute the index n for each term being calculated. Thus, \(T_2=3\times2-2=4\), \(T_3=4\times4-3=13\), and \(T_4=5\times13-4=61\). Therefore, the correct answer is \(61\). A close option such as \(59\) can result from an error in multiplying by 5 or subtracting 4 in the final step. Exam tip: To find \(T_4\), calculate \(T_2\) and \(T_3\) in order first.

A recursive definition needs an initial term and a rule to generate each next term. In A, \(a_1\) is given, so \(a_2=3+4=7\). B lacks an initial term, while D also needs \(a_2\). Exam tip: always check that all required starting terms are provided.

Using the recursive rule, \(a_3=a_2+3(3)=a_2+9\) and \(a_4=a_3+3(4)=a_3+12\). Hence, \(31=a_2+9+12=a_2+21\), so \(a_2=10\). If \(a_2=12\), then \(a_4\) would be 33, not 31. Exam tip: When a later term is given, expand the recurrence backwards to reach the required term.

Using the recursive rule, \(b_2=2b_1+2\). Hence, \(b_3=2b_2+3=2(2b_1+2)+3=4b_1+7\). Since \(b_3=19\), we get \(4b_1+7=19\), so \(b_1=3\). If 2 is used instead, it gives \(b_3=15\), so it is not correct. Exam tip: while expanding a recurrence, do not forget to add the value of the index at each step.

Using the recursive rule with \(n=2\), \(c_2=c_1+2^2=2+4=6\). Then, for \(n=3\), \(c_3=c_2+3^2=6+9=15\). Therefore, option D is correct. In option C, \(c_2\) is correct, but \(3^2=9\) has not been added correctly to find \(c_3\). Exam tip: to obtain each new term, add the square of the current index to the previous term.

(d_4=28) and (d_5=43), so the condition is first satisfied at (n=5). In exams, choose the first term that crosses the limit.

Using the recursive rule, \(e_2=2(3)+2=8\), \(e_3=2(8)+3=19\), \(e_4=2(19)+4=42\), and \(e_5=2(42)+5=89\). Therefore, \(e_5-e_4=89-42=47\). The value 49 does not result because finding \(e_5\) requires using \(2e_4\). Exam tip: In a recursive sequence, calculate the required terms one by one in order.

The first four terms are (2), (6), (12), and (20), and their sum is (40). In exams, add all terms when the sum is asked.

Using the recursive rule, \(g_2=3(1)-1=2\), \(g_3=3(2)-1=5\), and \(g_4=3(5)-1=14\). Hence, \(g_4:g_3=14:5\), so option B is correct. A ratio such as \(12:5\) results from using an incorrect term value. Exam tip: calculate the terms in order up to the required term and keep the ratio in the stated order \(g_4:g_3\).

Apply the recursive rule successively: \(h_3=2(7)-2+3=15\), \(h_4=2(15)-7+3=26\), and \(h_5=2(26)-15+3=40\). Therefore, the correct answer is 40. A value such as 38 can result from mishandling the signs of the previous terms or the \(+3\) term. Exam tip: before substituting, write down \(h_{n-1}\) and \(h_{n-2}\) separately for each new term.

In the recursive rule, add n^2-n to the preceding term for each value of n. Thus, p_2=1+(2^2-2)=3, p_3=3+(3^2-3)=9, p_4=9+(4^2-4)=21, and p_5=21+(5^2-5)=41. Therefore, the correct answer is 41. A value such as 43 can result from incorrectly handling the first increment, 2. Exam tip: To find p_5, list the successive values from n=2 to n=5 systematically.

(a_2=12), (a_3=29), (a_4=64), and (a_5=135). In exams, add the current value of (n) correctly.

The governing concept is a recursive rule: each term is calculated from the preceding term, and the value of n must be substituted separately at every step. Start with b₁ = 7. For n = 2, b₂ = 3(7) − 2(2) + 1 = 21 − 4 + 1 = 18. For n = 3, b₃ = 3(18) − 6 + 1 = 49. For n = 4, b₄ = 3(49) − 8 + 1 = 147 − 7 = 140. Thus option B is correct. Option A can result from an arithmetic slip in the final subtraction, while 144 and 148 do not follow the stated recurrence. It is important not to treat the expression as a constant-difference sequence.

Use the recursive rule step by step: \(c_3=5+2(2)+3=12\), \(c_4=12+2(5)+4=26\), and \(c_5=26+2(12)+5=55\). Therefore, the correct answer is \(55\). A close option such as \(53\) may result from not including the term \(2c_{n-2}\) correctly. Exam tip: write the previous two terms and the current value of \(n\) separately before substituting.