Recursive rule

The governing concept is a two-term recurrence: each new term depends on both preceding terms, so the second previous term must not be omitted. The initial values are a₁ = 2 and a₂ = 5. For n = 3, a₃ = 2a₂ + a₁ = 2(5) + 2 = 12. For n = 4, a₄ = 2a₃ + a₂ = 2(12) + 5 = 29. For n = 5, a₅ = 2a₄ + a₃ = 2(29) + 12 = 70. Therefore option C is correct. The sequence is 2, 5, 12, 29, 70. A result based only on 2aₙ₋₁ would violate the stated rule; arithmetic errors or using nonconsecutive terms can explain the other distractors.

This is a second-order recursive rule, so each new term uses the two preceding terms. Calculate a₃ = 3(7) − 2(3) = 21 − 6 = 15. Then a₄ = 3(15) − 2(7) = 45 − 14 = 31. Finally, a₅ = 3(31) − 2(15) = 93 − 30 = 63. Thus option C is correct; using only one previous term would not follow the stated rule.

Apply the recursive rule successively for each value of n. \(a_2=4+2^1+1^2=7\), \(a_3=7+2^2+2^2=15\), and \(a_4=15+2^3+3^2=15+8+9=32\). Therefore, the correct answer is 32. A value such as 29 can result from adding \(3^2\) incorrectly. Exam tip: to find \(a_4\), substitute \(n=1,2,3\) in order.

Using n=1,2,3,4 in the recursive rule, the values subtracted are 3, 6, 11, and 20. Hence, a_5=120-(3+6+11+20)=120-40=80. The value 82 is obtained after only the first three subtractions, so it is a_4, not a_5. Exam tip: to find a_5 from a_1, apply the rule four times.

The governing concept is recursive evaluation with an alternating sign. Each new term must be calculated from the immediately preceding term, and the exponent n + 1 determines whether the added quantity is positive or negative. Starting with a₁ = 10: for n = 1, a₂ = 10 + (−1)²(3) = 13; for n = 2, a₃ = 13 + (−1)³(4) = 9; for n = 3, a₄ = 9 + (−1)⁴(5) = 14; and for n = 4, a₅ = 14 + (−1)⁵(6) = 8. Therefore option B is correct. Ignoring the alternating signs or using the wrong index would produce a different value.

This rule builds the sequence by adding a quantity that depends on n. Since the required term is a₄ and the sequence starts at a₁, the rule must be used three times, with n equal to 1, 2, and 3. The added amount is always n(n + 2) + 1, so the multiplication and the final addition must both be performed at each step.

For n = 1, the added amount is 1(3) + 1 = 4, giving a₂ = 5 + 4 = 9. For n = 2, it is 2(4) + 1 = 9, giving a₃ = 18. For n = 3, it is 3(5) + 1 = 16, giving a₄ = 34. Thus the successive additions are 4, 9, and 16, whose sum is 29; 5 + 29 also equals 34. Therefore option C is correct. Keeping n matched to the step prevents using the wrong expression.

Use n = 1, 2, and 3 successively because these produce a₂, a₃, and a₄. The amounts subtracted are 1(3)+1 = 4, 2(4)+1 = 9, and 3(5)+1 = 16. Therefore a₂ = 150−4 = 146, a₃ = 146−9 = 137, and a₄ = 137−16 = 121. Option C is correct; subtracting only n(n+2) would omit the final −1.

Use the current value of n at each step of the recursive rule. a_2=2(2)+3(1)-1=6, a_3=2(6)+3(2)-1=17, and a_4=2(17)+3(3)-1=42. Therefore, 42 is correct. A value such as 39 can result from an error while calculating 2a_3 in the final step. Exam tip: explicitly write n=1, 2, and 3 while finding a_2, a_3, and a_4 respectively.

The recurrence is \(a_{n+1}=\frac{a_n}{2}+n^2\), with initial value \(a_1=64\). To find \(a_2\), use \(n=1\): \(a_2=\frac{64}{2}+1^2=32+1=33\). To find \(a_3\), use \(n=2\): \(a_3=\frac{33}{2}+2^2=\frac{33}{2}+4=\frac{33}{2}+\frac{8}{2}=\frac{41}{2}\). Hence option B is correct.

The index must change from 1 to 2 because the first application produces the second term and the second application produces the third term. Keeping the fraction exact avoids rounding and gives the precise result. The decimal value would be 20.5, but \(\frac{41}{2}\) is the exact form requested by the options. The other fractional choices differ by 1 from this result and arise from an arithmetic error in the final addition.

The governing concept is a recursive sequence: a new term depends on the immediately preceding term, so a₂ must be found before a₃. Begin with a₁ = 12. Substituting n = 1 gives a₂ = a₁/3 + 2(1) = 12/3 + 2 = 4 + 2 = 6. Now substitute n = 2 and use a₂ = 6: a₃ = a₂/3 + 2(2) = 6/3 + 4 = 2 + 4 = 6. Thus the value of a₃ is 6, so option C is correct. Option A comes from omitting 2n; other distractors reflect a wrong index or an arithmetic error. The recurrence must be followed in order.

The governing concept is an index-dependent second-order recurrence. Each term is obtained by adding the two preceding terms and then adding the current index n. Beginning with a₁ = 1 and a₂ = 4, use n = 3 to get a₃ = a₂ + a₁ + 3 = 4 + 1 + 3 = 8. Next, a₄ = a₃ + a₂ + 4 = 8 + 4 + 4 = 16. Finally, for n = 5, a₅ = a₄ + a₃ + 5 = 16 + 8 + 5 = 29. Therefore option C is correct. The added index cannot be omitted: without it the values would be smaller. Options A, B, and D can result from using the wrong preceding terms, forgetting one index contribution, or making an addition error.

In a recursive rule, each new term is calculated from the preceding term. First, \(a_2=2^2+2+1=7\). Then \(a_3=7^2+7+1=49+7+1=57\). Therefore, the correct answer is 57. A value such as 59 usually results from an error while squaring or adding. Exam tip: write the previous term separately before calculating the next term.

In a recursive rule, each new term is calculated from the preceding term. First, \(a_2=6^2-4(6)=36-24=12\). Then \(a_3=12^2-4(12)=144-48=96\). Therefore, the correct answer is 96. A value such as 84 can result from an error in the second calculation. Exam tip: always find \(a_2\) before calculating \(a_3\).

(53-9=44), and (44\div4=11) steps, so the term number is (12). Exam tip: add (1) to the number of steps.

For (a_{10}), (6) is added (9) times, so (7+54=61). Exam tip: the rule is applied one less time than the term number.

Using the recursive rule, \(a_2=2(3)+5=11\), \(a_3=2(11)+5=27\), and \(a_4=2(27)+5=59\). Therefore, \(a_4-a_2=59-11=48\), so option C is correct. A value such as \(45\) can result from an error in finding an intermediate term or subtracting. Exam tip: write the required terms in order before taking their difference.

The recursive rule says that each new term is the sum of the previous two terms plus 5. Thus, \(a_3=6+2+5=13\), \(a_4=13+6+5=24\), and \(a_5=24+13+5=42\). Therefore, 42 is correct. A value such as 46 results from using the preceding terms or the added constant incorrectly. Exam tip: at every step, include exactly the previous two terms and the +5.

In \(a_{n+1}=-2a_n\), the factor 2 doubles the magnitude and the minus sign reverses the sign. \(2a_n\) does not reverse it. Exam tip: check a multiplier’s sign and magnitude separately.

A recursive rule must reproduce the sequence from its first term. For option A, the initial value is \(a_1=2\), and the next-term rule is \(a_{n+1}=2a_n+n^2\). To test it, use \(n=1\) to produce the second term and \(n=2\) to produce the third term. Matching successive terms is the key check.

Using option A, \(a_2=2(2)+1^2=5\), and \(a_3=2(5)+2^2=14\). Continuing, \(a_4=2(14)+3^2=37\), exactly as required. Option B would give 11 for the third term, option C starts with the wrong first term, and option D gives 5 followed by 14? Actually it gives \(a_3=14\) but \(a_4=41\), so it fails later. Therefore A is correct.

A recursive rule must specify the correct starting term and reproduce every successive term from the previous one. The sequence begins with 6, and its successive differences are 9−6 = 3, 14−9 = 5, and 21−14 = 7. In the rule aₙ₊₁ = aₙ + 2n + 1, the added amounts are 3 for n = 1, 5 for n = 2, and 7 for n = 3. Starting from a₁ = 6, it therefore produces 9, 14, and 21 exactly. Thus option B is correct. Option A adds 2, 4, and 6, so it produces a different sequence. Option C has the wrong first term, while option D produces 6, 9, 15, 27. Checking both the initial value and the transition confirms the answer.