Recursive rule

This is a recursive sequence with an alternating-sign increment. At each step, evaluate (−1)ⁿ first, then multiply by n. Since s₁ = 3, for n = 2 we have s₂ = 3 + (+1)·2 = 5. For n = 3, s₃ = 5 + (−1)·3 = 2. For n = 4, s₄ = 2 + (+1)·4 = 6. Finally, for n = 5, s₅ = 6 + (−1)·5 = 1. Therefore option D is correct. The important point is that even indices contribute positive amounts and odd indices contribute negative amounts here. Confusing the sign pattern, beginning the recurrence at n = 1, or treating every increment as positive can produce the distractor values.

Using the recursive rule, \(t_2=x+2\), \(t_3=x+2+3\), \(t_4=x+2+3+4\), and \(t_5=x+2+3+4+5=x+14\). Since \(t_5=24\), we get \(x+14=24\), so \(x=10\). If \(x=11\), then \(t_5=25\), so it is not correct. Exam tip: In a recursive sequence, list every added term up to the required term before solving for the first term.

The correct rule is \(a_1=2\) and \(a_n=3a_{n-1}\), because every term is three times the preceding term: \(2\times3=6\), \(6\times3=18\), and \(18\times3=54\). Option B gives the second term as 6, but its next term would be \(2\times6+2=14\), not 18. Exam tip: when ratios of consecutive terms are equal, check for a multiplicative recursive rule.

In the recursive rule, multiply the preceding term by 2 and then subtract 1. Thus, \(v_2=2(4)-1=7\), \(v_3=2(7)-1=13\), \(v_4=2(13)-1=25\), and \(v_5=2(25)-1=49\). Therefore, the correct answer is 49. Although 47 is a close distractor, it is not obtained by applying the given rule successively. Exam tip: at each step, double the previous term before subtracting 1.

Apply the recursive rule step by step: \(w_2=0+2^1=2\), \(w_3=2+2^2=6\), \(w_4=6+2^3=14\), and \(w_5=14+2^4=30\). Therefore, the correct answer is 30. A value such as 28 can result from using the final increment incorrectly. Exam tip: from \(w_1\) to \(w_5\), apply the rule four times.

In a recursive rule, first find the preceding term. \(x_2=2x_1+2^2=2(1)+4=6\). Then \(x_3=2x_2+3^2=2(6)+9=21\). Therefore, the correct answer is 21. A value such as 18 can result from incorrectly using \(x_1\) in place of \(x_2\). Exam tip: substitute the immediately preceding term at each step.

Apply the recursive rule step by step: (y_2=7+1^2=8), (y_3=8+2^2=12), (y_4=12+3^2=21), and (y_5=21+4^2=37). Therefore, the correct answer is 37. A value such as 39 can result from adding the squares or the initial term incorrectly. Exam tip: to find (y_5), include exactly the increments (1^2,2^2,3^2,4^2).

By the recursive rule, each new term is obtained by multiplying the previous two terms and adding 1. Thus, z_3=3×2+1=7, z_4=7×3+1=22, and z_5=22×7+1=155. Therefore, the correct answer is 155. A value such as 153 can result from forgetting the final “+1” in the rule. Exam tip: write each term in sequence and use both preceding terms at every step.

A recursive rule uses the preceding term to find each new term. \(A_2=2\times1+1=3\), \(A_3=3\times3+1=10\), and \(A_4=4\times10+1=41\). Hence, the correct answer is 41. An option such as 37 results from not applying the multiplication by 4 correctly while finding \(A_4\). Exam tip: write the value of \(n\) first, then multiply the previous term by it and add 1.

In the recursive rule, each term is obtained by adding the relevant power to the preceding term. Thus, \(B_2=9+2^2=13\), \(B_3=13+2^3=21\), and \(B_4=21+2^4=37\). Therefore, 37 is correct. Getting 39 would result from using an incorrect exponent or starting the additions incorrectly. Exam tip: when \(B_1\) is given, use \(n=2\) in the rule to find the next term.

Apply the recursive rule step by step: \(C_2=5+2(2-1)=7\), \(C_3=7+3(3-1)=13\), and \(C_4=13+4(4-1)=25\). Hence, the correct answer is \(25\). A close distractor such as \(27\) can result from adding \(4(4-1)=12\) incorrectly. Exam tip: calculate \(n(n-1)\) first for each term, then add it to the preceding term.

Each term is found from the previous term using the recursive rule. Thus, \(D_2=2\times2+1=5\), \(D_3=2\times5+2=12\), and \(D_4=2\times12+3=27\). Therefore, the correct answer is 27. An answer such as 25 can result from adding the \(n-1\) term incorrectly. Exam tip: To find \(D_4\), calculate \(D_2\) and \(D_3\) in order first.

In the recursive rule, each new term is obtained by adding 2 to the sum of the two immediately preceding terms. Thus, E_3=1+4+2=7, E_4=4+7+2=13, and E_5=7+13+2=22. Therefore, the correct answer is 22. The value 20 can result from not using the correct two previous terms in the final step. Exam tip: Write the terms one by one, starting from E_3.

The governing idea is recursive evaluation with an index-dependent decrement. Rewrite the rule as Fₙ = Fₙ₋₁ − (2n − 1), showing that the successive amounts subtracted are odd numbers. Starting with F₁ = 20: F₂ = 20 − 4 + 1 = 17; F₃ = 17 − 6 + 1 = 12; F₄ = 12 − 8 + 1 = 5; and F₅ = 5 − 10 + 1 = −4. Hence option B is correct. Equivalently, the total subtraction is 3 + 5 + 7 + 9 = 24, so 20 − 24 = −4. Confusing the starting index, dropping +1, or treating the changes as additions leads to the other options.

In a recursive rule, each new term is found from the preceding term. Thus, \(G_2=2+2(3)=8\), \(G_3=3+2(8)=19\), and \(G_4=4+2(19)=42\). Therefore, the correct answer is 42. A value such as 44 can result from using an incorrect previous term. Exam tip: to find \(G_4\), calculate \(G_2\) and \(G_3\) in order first.

In a recursive rule, each term is found from the previous term. \(H_2=2+2(3)=8\), \(H_3=8+3(4)=20\), and \(H_4=20+4(5)=40\). Hence, the correct answer is \(40\). A value such as \(38\) can result from using an incorrect increase instead of \(4(5)\) in the final step. Exam tip: Start with \(H_1\) and calculate the terms successively for \(n=2,3,4\).

Apply the recursive rule using the current value of n: J_2=50-3(1)=47, J_3=47-3(2)=41, J_4=41-3(3)=32, and J_5=32-3(4)=20. Therefore, the correct answer is 20. The value 23 may result from subtracting only 3 in the last step instead of 3(n-1), which does not follow the rule. Exam tip: To move from J_1 to J_5, write the four steps for n=2,3,4,5.

(K_3=3), (K_4=7), and (K_5=17). In exams, double the previous term and add the term before it.

Each new term is obtained by adding \(2n+1\) to the preceding term. Thus, \(L_2=6+5=11\), \(L_3=11+7=18\), \(L_4=18+9=27\), and \(L_5=27+11=38\). Therefore, 38 is correct. A value such as 40 results from using an incorrect increment at one of the steps. Exam tip: in a recursive rule, substitute consecutive values of \(n\), starting with the second term.

For a recursive rule, first calculate the preceding term. \(M_2=3M_1+2=3\times1+2=5\). Then \(M_3=3M_2+3=3\times5+3=18\). Therefore, 18 is correct. An answer such as 16 can result from using an incorrect value of \(M_2\) or of \(n\). Exam tip: remember to add the current index \(n\) at every step.