Recursive rule

The recursive rule forms each new term by adding the previous two terms and 6. Thus, \(a_3=8+4+6=18\), and \(a_4=18+8+6=32\). Therefore, the correct answer is 32. The value 26 is only \(a_2+a_3\); it misses the required addition of 6. Exam tip: find \(a_3\) before calculating \(a_4\), and include the constant at every step.

In an arithmetic progression, the difference between consecutive terms is constant. Here, \(a_{n+1}-a_n=4\), so every term increases by 4. In option C, the difference is \(4n\), which changes. Exam tip: test an AP rule using \(a_{n+1}-a_n\).

Apply the recursive rule successively for n=1, 2, and 3: a_2=5+(8×1-3)=10, a_3=10+(8×2-3)=23, and a_4=23+(8×3-3)=44. Therefore, the correct answer is 44. A value such as 41 results from using an incorrect increment at one of the steps. Exam tip: to find a_4, apply the rule step by step from n=1 to n=3.

Putting n=1 gives a_2=a_1+8(1)-3=5+5=10. Then, putting n=2 gives a_3=a_2+8(2)-3=10+13=23. Therefore, the correct answer is 23. A value such as 22 may result from incorrectly treating the increment as constant, whereas 8n-3 changes with n. Exam tip: in a recursive relation, substitute the correct value of n at every step.

The original options did not contain the mathematically correct value, so the options have been corrected. Apply the recursive rule for n = 1, 2, and 3 because a₄ requires three transitions. First, a₂ = 180 − (6×1 + 5) = 180 − 11 = 169. Next, a₃ = 169 − (6×2 + 5) = 169 − 17 = 152. Finally, a₄ = 152 − (6×3 + 5) = 152 − 23 = 129. Therefore option A, 129, is correct. The subtractions are 11, 17, and 23, increasing by 6. The previously supplied value 136 would result from an arithmetic or indexing mistake and is not compatible with the stated recurrence.

Using the recursive rule, first put n=1: a_2=180-(6×1+5)=169. Next, put n=2: a_3=169-(6×2+5)=169-17=152. Therefore, 152 is correct. The value 150 would result from incorrectly subtracting 19 in the second step. Exam tip: remember that the value of n changes to 1, 2, 3, and so on for successive terms.

First, use the recursive rule to get \(a_3=4a_2-3a_1+2=4(4)-3(2)+2=12\). Then \(a_4=4a_3-3a_2+2=4(12)-3(4)+2=38\). Therefore, 38 is the correct answer. A value such as 36 can result from forgetting the constant term \(+2\). Exam tip: write the two preceding terms separately before applying their coefficients.

In \(a_n=a_{n-1}+a_{n-2}\), every new term depends on the two preceding terms, so both \(a_1\) and \(a_2\) are needed. Exam tip: count the previous terms used to identify the required initial terms.

The direct answer is option B, 31. Starting from \(a_1=6\), reaching \(a_5\) requires four updates. The increments for n=1,2,3,4 are \(\frac52,5,\frac{15}2,10\). Their sum is \(\frac{5(1+2+3+4)}2=\frac{5\cdot10}{2}=25\). Therefore \(a_5=6+25=31\). Option B is correct. Option A, 29, is two less than the result; option C, 33, is two more; and option D, 35, is four more. The fraction applies to the whole product \(5n\), so calculate each increment carefully. A common mistake is to use only the first three increments, which would reach \(a_4\), not \(a_5\). Memory cue: the number of transitions from term 1 to term 5 is four.

Direct answer: C, 35. Moving from a₁ to a₅ requires four transitions, so use n=1,2,3,4. The recurrence subtracts 5n/2 at each transition. The subtracted amounts are 5/2, 5, 15/2, and 10. Their total is (5/2)(1+2+3+4)=(5/2)(10)=25. Hence a₅=60−25=35. A term-by-term check confirms this: a₂=60−5/2=57.5, a₃=57.5−5=52.5, a₄=52.5−15/2=45, and a₅=45−10=35. Thus C is correct. A, 30, subtracts too much. B, 33, does not follow from the four required decrements. C, 35, uses every transition correctly. D, 38, usually results from omitting one part of the decrement or making a fraction error. Remember that the journey from the first term to the fifth term has four steps, not five.

In a recursive rule, substitute the current index value at each step. For \(n=1\), \(a_2=1\times3+4=7\). Then, for \(n=2\), \(a_3=2\times7+4=18\), and for \(n=3\), \(a_4=3\times18+4=58\). Therefore, 58 is correct. An answer such as 52 may result from not changing \(n\) correctly at every step. Exam tip: write down the value of \(n\) before calculating each next term.

In an AP, the difference between consecutive terms is constant. In A, \(a_{n+1}-a_n=-5\) for every \(n\), so the common difference is \(-5\). In C, the difference \(-n\) changes. Exam tip: first check \(a_{n+1}-a_n\).

The governing concept is a recursive rule, which specifies an initial term and an operation for obtaining each following term from the preceding term. To verify a rule, start with the stated first term and apply the recurrence repeatedly; it must reproduce all displayed terms. For option A, a₁ = 5. Then a₂ = 3(5) + 1 = 16, a₃ = 3(16) + 1 = 49, and a₄ = 3(49) + 1 = 148. Thus it matches the entire sequence. Option B produces 16 first but then 2(16) + 6 = 38, not 49. Option C begins with the wrong initial term, and option D gives 14 as the second term. Therefore A is the only correct rule.

Direct answer: B, the fifth term. Start with a₁=7 and apply the recurrence successively. For n=1, a₂=7+1²+1=9. For n=2, a₃=9+2²+2=15. For n=3, a₄=15+3²+3=27. For n=4, a₅=27+4²+4=47. Therefore 47 is the fifth term, so B is correct. A is wrong because the fourth term is 27, not 47. B correctly identifies both the value and its position. C and D refer to later terms; they would require continuing the recurrence beyond a₅. The key distinction is between a term’s value and its position: 47 is the value, while 5 is the index. A common mistake is to count the number of additions rather than include the starting term; beginning at a₁ means that after four transitions we reach a₅.

The direct answer is option B, the fifth term. The recurrence is \(a_{n+1}=a_n-(n^2+n)\), with \(a_1=150\). To find the terms, use \(n=1\) to obtain \(a_2\): subtract \(1^2+1=2\), so \(a_2=150-2=148\). Next use \(n=2\): subtract \(2^2+2=6\), so \(a_3=148-6=142\). Next use \(n=3\): subtract \(3^2+3=12\), so \(a_4=142-12=130\). Next use \(n=4\): subtract \(4^2+4=20\), so \(a_5=130-20=110\). Thus 110 is the fifth term. Option A, fourth, is wrong because the fourth term is 130. Option B, fifth, is correct. Option C, sixth, would require one more subtraction and would give 80. Option D, seventh, would require two more steps and would not be 110. Always write the index beside each term so that the position is not confused with the value.

Apply the recursive rule using the two preceding terms. \(a_3=8+2(2)+3=15\), \(a_4=15+2(8)+4=35\), and \(a_5=35+2(15)+5=70\). Therefore, the correct answer is 70. A value such as 66 can result from an error in calculating the \(2a_{n-2}\) term. Exam tip: write the current value of \(n\) separately at every step.

The rule \(a_n=3a_{n-1}-2a_{n-2}\) depends on the two preceding terms, so both \(a_1\) and \(a_2\) are needed. A and B are first-order rules and need only one initial term. Exam tip: count the previous terms used in the rule.

In the recursive rule, the fixed first term \(a_1=6\) and the relevant \(n^2\) are added at each step. Thus, \(a_2=6+6+1^2=13\), \(a_3=13+6+2^2=23\), and \(a_4=23+6+3^2=38\). Therefore, the correct answer is \(38\). A close option such as \(37\) may result from evaluating \(3^2\) incorrectly. Exam tip: to find \(a_4\), use \(n=3\) in the final recurrence step, not \(n=4\).

Use the recursive rule step by step. For \(n=1\), \(a_2=a_1+a_1+1^2=6+6+1=13\). Then, for \(n=2\), \(a_3=a_2+a_1+2^2=13+6+4=23\). Therefore, the correct answer is 23. The value of \(a_1\) remains 6 in every step; replacing it with \(a_2\) would be incorrect. Exam tip: To find \(a_3\), apply the rule first with \(n=1\) and then with \(n=2\).

The added values (2!,3!,4!) are (2,6,24), so (a_4=37). Exam tip: calculate factorial increments in order.

Direct answer: a₅ = 110, so option C is correct. This is a recursive rule, meaning each new term is calculated from the previous term. To find a₅, apply the rule four times, because we begin with a₁ and move successively to a₂, a₃, a₄, and a₅. The correction term is (−1)ⁿn. When n is odd, (−1)ⁿ = −1; when n is even, (−1)ⁿ = +1. Thus a₂ = 2(7) − 1 = 13, a₃ = 2(13) + 2 = 28, a₄ = 2(28) − 3 = 53, and a₅ = 2(53) + 4 = 110. Option A, 104, is not the result of the four correct recursive steps. Option B, 108, also comes from an incorrect sign or arithmetic step. Option C, 110, follows every step and is therefore correct. Option D, 114, usually results from adding the correction with the wrong value or sign. Memory cue: odd n means subtract, even n means add; always use the n belonging to the transition being calculated.

This is a recursive sequence because each term is obtained from the preceding term. For n = 1, subtract 3¹ + 1 = 4, giving a₂ = 196. For n = 2, subtract 3² + 2 = 11, giving a₃ = 185. For n = 3, subtract 3³ + 3 = 30, so a₄ = 185 − 30 = 155. Therefore, option A is correct; the other values result from an arithmetic error in one subtraction.

The given rule uses the two preceding terms, so calculate successively. First, a₃ = 3a₂ − 2a₁ = 3(11) − 2(5) = 23. Next, a₄ = 3(23) − 2(11) = 47. Finally, a₅ = 3(47) − 2(23) = 141 − 46 = 95. Thus option C is correct. The nearby alternatives usually come from using only one earlier term or making an arithmetic error.

Apply the recurrence for n = 1, 2, and 3. The first decrement is 1(3) + 2¹ = 5, so a₂ = 180 − 5 = 175. The second is 2(4) + 2² = 12, so a₃ = 163. The third is 3(5) + 2³ = 23, so a₄ = 163 − 23 = 140. Hence option C is correct; the other choices reflect an incorrect value of one decrement.

Direct answer: a₅ = 61, so option D is correct. The rule is recursive and uses two preceding terms, not just one. It also adds 3n, where n is the index of the term currently being found. Begin with the given values a₁ = 2 and a₂ = 4. For n = 3, a₃ = a₂ + a₁ + 3(3) = 4 + 2 + 9 = 15. For n = 4, a₄ = a₃ + a₂ + 3(4) = 15 + 4 + 12 = 31. For n = 5, a₅ = a₄ + a₃ + 3(5) = 31 + 15 + 15 = 61. Option A, 52, does not include the correct preceding terms and index. Option B, 55, reflects an arithmetic or indexing error. Option C, 58, may result from using only part of the recurrence. Option D, 61, includes both earlier terms and the correct 3×5 contribution. Common warning: in a two-term recurrence, never omit either aₙ₋₁ or aₙ₋₂, and use the current n in 3n.