Recursive rule

The governing concept is a recursive sequence evaluated backward. The recurrence adds 5n − 1 when moving from dₙ₋₁ to dₙ. Thus the increments are 9 for n = 2, 14 for n = 3, 19 for n = 4 and 24 for n = 5. Starting from d₁, d₅ = d₁ + 9 + 14 + 19 + 24 = d₁ + 66. Since d₅ = 79, d₁ = 79 − 66 = 13. Therefore option A is correct. A useful check is to rebuild the sequence forward: 13 + 9 = 22, then +14 = 36, then +19 = 55, then +24 = 79. The other choices arise if one increment is omitted, calculated with the wrong n, or subtracted incorrectly.

Each new term is obtained by doubling the previous term and adding (2). In exams, test the rule at two consecutive places.

(f_5=109) and (f_6=221), so the condition is first satisfied at (n=6). In exams, choose the first term that crosses the limit.

The governing concept is a recursive rule with an alternating sign. At each step, calculate (−1)ⁿ first: it is positive for even n and negative for odd n. Starting with g₁ = 10, for n = 2 the added amount is (+1)(2²) = 4, so g₂ = 14. For n = 3 the amount is (−1)(3²) = −9, so g₃ = 5. For n = 4, add 16 to obtain g₄ = 21. For n = 5, subtract 25, giving g₅ = 21 − 25 = −4. Hence option C is correct. The distractors reflect sign errors or failure to square n; the alternating factor must be applied before adding each term.

In a recursive rule, each term is calculated from the preceding term. First, \(h_2=(2-1)\times2+3=5\). Then \(h_3=(3-1)\times5+3=13\), and \(h_4=(4-1)\times13+3=42\). Therefore, the correct answer is 42. A value such as 45 may result from incorrectly using \(n\) instead of \(n-1\), or from using the wrong multiplier. Exam tip: substitute the value of \(n\) first and then use the previous term carefully.

In a first-order rule, each new term depends only on the preceding term. Thus, one initial term plus the rule fixes every later term. A rule alone can produce many sequences. Exam tip: the number of initial values matches the order of recurrence.

\(k_2=42\), \(k_3=24\), \(k_4=16\), and \(k_5=13\). In exams, add (n) after halving the previous term.

Apply the recursive rule step by step: \(m_2=3(2)+2-2=6\), \(m_3=3(6)+3-2=19\), and \(m_4=3(19)+4-2=59\). Hence, the correct value is \(59\). A close option such as \(61\) may result from adding the \(n-2\) term incorrectly. Exam tip: write the value of \(n\) separately at every step before substituting.

Apply the recursive rule using the preceding term each time: \(N_2=2(8)-2(1)=14\), \(N_3=2(14)-3(2)=22\), and \(N_4=2(22)-4(3)=32\). Hence, the correct answer is \(32\). A close distractor such as \(34\) can result from evaluating \(4(3)\) incorrectly. Exam tip: calculate \(n(n-1)\) separately at every step before subtracting it.

(r_3=11), (r_4=20), and (r_5=34). In exams, do not change the order of the older term being subtracted.

The governing concept is recursive evaluation with alternating additions and subtractions of consecutive odd numbers. Begin with s₁ = 5. For n = 2, (−1)³ = −1 and 2n − 1 = 3, so s₂ = 5 − 3 = 2. For n = 3, the sign is positive and the amount is 5, giving s₃ = 7. For n = 4, subtract 7 to obtain s₄ = 0. For n = 5, add 9, so s₅ = 9. Therefore option D is correct. The sequence of changes is −3, +5, −7, +9; confusing the exponent n + 1 or treating every change as positive leads to the other options. Writing the sign and magnitude separately prevents that error.

Using the recursive rule, t_2=2x+3, t_3=2(2x+3)+3=4x+9, and t_4=2(4x+9)+3=8x+21. Since t_4=61, we get 8x+21=61, so 8x=40 and x=5. If x were 6, then t_4 would be 69, not 61. Exam tip: Expand the recurrence step by step up to the given term before solving.

Each new term is obtained by doubling the previous term and adding (3). In exams, check the rule, not only the difference.

In the recursive rule, each new term is obtained by adding the cube of the current index to the previous term. Thus, (v_2=3+2^3=11), (v_3=11+3^3=38), and (v_4=38+4^3=38+64=102). Therefore, 102 is correct. An answer such as 108 may result from an error while adding the cubes or using the previous term. Exam tip: Start from n=2 after the initial term and calculate each term in order.

The recursive rule adds \(3^{n-1}+1\) to the preceding term at each step. Thus, \(w_2=1+3^1+1=5\), \(w_3=5+3^2+1=15\), and \(w_4=15+3^3+1=43\). Therefore, the correct answer is 43. A value such as 41 may result from forgetting to add \(+1\) in the final step. Exam tip: substitute each value of \(n\) separately and calculate the terms in order.

Apply the recursive rule successively for n=2, 3, and 4. x_2=2(4)+2^2-2=10, x_3=2(10)+3^2-3=26, and x_4=2(26)+4^2-4=64. Hence, the correct answer is 64. A value such as 68 can result from evaluating 4^2-4 incorrectly in the final step. Exam tip: substitute the value of n first and simplify the added term separately at every step.

Given \(y_1=2\), we get \(y_2=2+(2\times2-1)^2=2+9=11\), \(y_3=11+(2\times3-1)^2=11+25=36\), and \(y_4=36+(2\times4-1)^2=36+49=85\). Hence, the correct answer is eighty-five. Eighty-one can result from not applying the square correctly in the final step. Exam tip: in a recursive sequence, substitute each value of \(n\) step by step.

Under the recursive rule, each new term is found by multiplying the previous two terms and adding its term number. Thus, \(z_3=(3\times1)+3=6\), \(z_4=(6\times3)+4=22\), and \(z_5=(22\times6)+5=137\). Therefore, the correct answer is \(137\). A value such as \(145\) can result from using an incorrect previous term or term number. Exam tip: write the two preceding terms at every step before multiplying and adding the current \(n\).

Using the recursive rule, first put \(n=2\): \(A_2=(2+1)\times 2+2=8\). Then put \(n=3\): \(A_3=(3+1)\times 8+2=34\). Therefore, the correct answer is 34. The option 32 may result from forgetting the final \(+2\). Exam tip: write the value of \(n\) separately at each step before applying the multiplier.

Apply the recursive rule step by step: \(B_2=11+(2^2-2)=13\), \(B_3=13+(2^3-3)=18\), and \(B_4=18+(2^4-4)=30\). Hence, the correct answer is 30. A value such as 32 can result from calculating \(2^n-n\) incorrectly at one of the steps. Exam tip: find the terms successively for \(n=2,3,4\) rather than jumping directly from \(B_1\) to \(B_4\).