Recursive rule

Using the recursive rule successively for \(n=1,2,3,4\), we get \(a_2=3+1^3=4\), \(a_3=4+2^3=12\), \(a_4=12+3^3=39\), and \(a_5=39+4^3=103\). Hence, the correct answer is \(103\). The value \(106\) would result from using an incorrect final increment instead of \(4^3\). Exam tip: to find \(a_5\), add only the increments from \(1^3\) through \(4^3\).

Using n=1, 2, 3 and 4, the terms decrease by 1, 8, 27 and 64 respectively. Thus, a_5=150-(1+8+27+64)=150-100=50. Option 53 results from subtracting only the first three cubes; 4^3 must also be subtracted. Exam tip: to find a_5 from a_1, apply the recursive rule four times.

A recursive rule must reproduce later terms from the stated starting terms. Test option D using a₁ = 3 and a₂ = 8. For n = 3, a₃ = 2a₂ + a₁ = 2(8) + 3 = 19. For n = 4, a₄ = 2a₃ + a₂ = 2(19) + 8 = 46. Thus the rule generates both given terms and option D is correct. Option A gives a₃ = 8 + 2(3) = 14, so it fails immediately. Option B uses the wrong initial values, even though its expression resembles the required pattern. Option C produces a₃ = 8 + 3 + 8 = 19, but then produces a₄ = 19 + 8 + 8 = 35, not 46. Testing more than one step ensures the rule is genuinely correct.

Apply the recursive rule first for a₃: a₃ = a₂ + 3a₁ + 3 = 6 + 3×4 + 3 = 21. Then, a₄ = a₃ + 3a₂ + 4 = 21 + 3×6 + 4 = 43. Therefore, 43 is correct. A value such as 46 can result from using an incorrect previous term or index value. Exam tip: add the current value of n at every step.

In a recursive rule, substitute the correct value of n at each step. For n=1, a_2=2a_1+1=2×2+1=5. Then, for n=2, a_3=3a_2+2=3×5+2=17, and for n=3, a_4=4a_3+3=4×17+3=71. Therefore, the correct answer is 71. A value such as 68 can result from using the added term incorrectly in the final step. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

First (a_2=7), then (a_3=15), and finally (a_4=31). In exams, apply the recursive rule step by step.

Under the recursive rule, each new term is the previous term plus twice the term before it. Thus, t_3=5+2(2)=9, t_4=9+2(5)=19, and t_5=19+2(9)=37. Therefore, 37 is correct. A value such as 35 can result from using the wrong earlier terms or missing the coefficient 2. Exam tip: write down t_{n-1} and t_{n-2} separately before substituting their values.

In a recursive rule, use the index of the term currently being found. \(a_2=2(3)+2+1=9\), \(a_3=2(9)+3+1=22\), and \(a_4=2(22)+4+1=49\). Therefore, the correct answer is \(49\). An answer such as \(46\) can result from using an incorrect value of the index in the \(n+1\) part. Exam tip: write the new value of \(n\) at every step before calculating.

The governing concept is a recursive sequence: each term is calculated from the immediately preceding term, and n is substituted for the position of the term being found. Starting with b₁ = 6, calculate b₂ = 3(6) − 2 = 16. Then b₃ = 3(16) − 3 = 45. Finally, b₄ = 3(45) − 4 = 135 − 4 = 131. Thus option B is correct. The nearby alternatives can result from an arithmetic slip, such as forgetting to subtract the current index, using the wrong preceding term, or applying the subtraction before multiplication incorrectly. The recursive rule must be followed in order.

By the recursive rule, each new term equals twice the preceding term plus the term before it. Thus, \(c_3=2(3)+1=7\), \(c_4=2(7)+3=17\), and \(c_5=2(17)+7=41\). Therefore, the correct answer is 41. A close distractor such as 39 can result from using an incorrect previous term. Exam tip: write the two preceding terms at every step before calculating the next one.

Work backward using the recursive rule: \(d_4=d_3+8\), \(d_3=d_2+6\), and \(d_2=d_1+4\). Hence, \(d_4=d_1+4+6+8=d_1+18\). Since \(d_4=25\), \(d_1=25-18=7\). If \(d_1=8\), then \(d_4\) would be 26, not 25. Exam tip: when a later term is given in a recursive sequence, apply the rule step by step backward.

Each new term is obtained by doubling the previous term and adding (1). In exams, test the rule on two terms.

(f_2=8), (f_3=5), (f_4=1), and (f_5=-4). In exams, find the first term that goes below zero.

Adding (3), (5), (7), and (9) gives (g_5=26). In exams, write the added quantities separately.

Under the recursive rule, each term is found by multiplying the previous term by its index. Thus, \(h_2=2\times3=6\), \(h_3=3\times6=18\), and \(h_4=4\times18=72\). Therefore, 72 is correct. An option such as 64 does not follow the required successive multiplications by 2, 3, and 4. Exam tip: write the steps for \(n=2,3,4\) separately to avoid using the wrong index.

In the recursive rule, each term is obtained by adding the square of the current index to the preceding term. Thus, \(j_2=1+2^2=5\), \(j_3=5+3^2=14\), and \(j_4=14+4^2=30\). Therefore, the correct answer is 30. Adding \(1^2\) separately would be incorrect because \(j_1=1\) is already the given initial term. Exam tip: compute recursive sequences step by step from the initial term.

In a recursive rule, each new term is calculated from the preceding term. \(k_2=\frac{64}{2}+2=34\), \(k_3=\frac{34}{2}+3=20\), and \(k_4=\frac{20}{2}+4=14\). Hence, the correct answer is \(14\). A choice such as \(12\) can result from missing the division by 2 before adding 4. Exam tip: at every step, first divide the previous term by 2 and then add the current index \(n\).

In the recursive rule, substitute the current index each time. Thus, \(m_2=2+3(2)=8\), \(m_3=8+3(3)=17\), and \(m_4=17+3(4)=29\). Therefore, 29 is correct. A value such as 32 would result from using an incorrect increment in the final step instead of \(3n\). Exam tip: when \(m_1\) is given, begin finding the next term with \(n=2\).

The recurrence is \(n_n=2n_{n-1}-n^2\), with \(n_1=5\). To find the second term, substitute \(n=2\): \(n_2=2n_1-2^2=2(5)-4=6\). For the third term, substitute \(n=3\): \(n_3=2n_2-3^2=2(6)-9=3\). For the fourth term, substitute \(n=4\): \(n_4=2n_3-4^2=2(3)-16=6-16=-10\). Thus option C is correct.

The symbol on the left identifies the term being calculated, while the index in \(n^2\) changes at every step. One must not use the same square throughout the recurrence. The sequence begins \(5,6,3,-10\), so a negative fourth term is expected and is not a reason to stop. Option C, negative ten, matches the exact calculation; the other choices result from an error in the subtraction or in selecting the index.

By the recursive rule, each new term equals the previous term plus 3 times the term two places before it. Thus, \(r_3=2+3(1)=5\), \(r_4=5+3(2)=11\), and \(r_5=11+3(5)=26\). Therefore, the correct answer is 26. A value such as 23 can result from not multiplying the second previous term by 3 correctly. Exam tip: list the terms in order and identify \(r_{n-1}\) and \(r_{n-2}\) separately at each step.