Recursive rule

The governing concept is a recursive sequence in which every term depends on the two immediately preceding terms. Its rule is a_n=a_(n−1)+a_(n−2), with a_1=8 and a_2=13. Apply the rule one step at a time: a_3=8+13=21, a_4=13+21=34, a_5=21+34=55, and a_6=34+55=89. Thus option C is correct. The calculation must be continued successively; adding the first two values only once does not produce the sixth term. Option A is not obtained at any required step, and options B and D do not equal the sum of the fifth and fourth terms. The initial values and recurrence together determine a unique sequence.

The governing concept is a recursive rule: each term is generated from the two immediately preceding terms rather than from a single fixed difference or ratio. The given terms are \(a_1=2,a_2=3,a_3=5,a_4=8,a_5=13\). Continue carefully: \(a_6=a_4+a_5=8+13=21\), \(a_7=a_5+a_6=13+21=34\), and \(a_8=a_6+a_7=21+34=55\). Therefore option C is correct. Option A is the sixth term, and option B is the seventh term, so both result from stopping too early. Option D is the ninth term, obtained after one more addition, which shows why the term number must be tracked at every step.

The governing concept is a recursive addition rule. Starting with the fifth term 11, the next added number is 5, so the sixth term is \(11+5=16\). Add 6 to obtain the seventh term: \(16+6=22\). Add 7 to obtain the eighth term: \(22+7=29\). Therefore option D is correct. A useful equivalent check is \(a_8=1+(1+2+3+4+5+6+7)=1+28=29\). Option B is the seventh term, so it stops one step early. Options A and C do not result from applying the successive additions in order and therefore do not satisfy the stated recursive rule.

The governing idea is to inspect the successive differences rather than assume that the sequence has a constant common difference. The listed differences are -3, -5, -7, and -9; each new difference is 2 smaller than the preceding one. Therefore, the next difference must be -11. The last known term is -9, so the next term is obtained by adding this next difference: -9 + (-11) = -20. Hence option D is correct. Option C would result from using -9 again or making an arithmetic mistake, while the positive options ignore that the sequence is continuing downward. The pattern of odd negative differences confirms the result.

The governing concept is a recursive sequence: each term is calculated from the term immediately before it. The rule is next term = 2 × previous term + 1. Using the last given term, 2 × 127 + 1 = 254 + 1 = 255, so option C is correct. Options A and B result from incorrect arithmetic, while D does not follow the stated rule.

The governing concept is a recursive rule, where the next value depends on the preceding value. Here, next term = 2 × previous term − 1. Applying it to the final term 33 gives 2 × 33 − 1 = 66 − 1 = 65. Therefore option C is correct. Option A would subtract 3, option B omits the subtraction, and option D uses neither the correct operation nor result.

The rule \(a_{n+1}=a_n+2n+1\) must be applied with the current value of \(n\) at each step. Start with \(a_1=3\). For \(n=1\), \(a_2=3+2(1)+1=6\). For \(n=2\), \(a_3=6+2(2)+1=11\). For \(n=3\), \(a_4=11+2(3)+1=18\). Finally, for \(n=4\), \(a_5=18+2(4)+1=27\). Thus option C is correct.

A common mistake is to use the same value of \(n\) repeatedly or to stop after calculating \(a_4\). The index in the recurrence identifies the step being performed, so it changes from 1 to 4. The resulting terms are \(3,6,11,18,27\), and the fifth term is therefore 27, not any of the other listed values.

In the recursive rule, the value of n changes at each step. Thus, a_2=40-3(1)=37, a_3=37-3(2)=31, and a_4=31-3(3)=22. Therefore, the correct answer is 22. Choosing 16 would result from incorrectly using a fixed subtraction instead of the changing value of 3n. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

This is a recursively defined sequence because each term is obtained from the preceding term rather than directly from its position. Begin with a₁ = 2 and apply the rule using the current value of n. For n = 1, a₂ = 2a₁ + 1 = 2(2) + 1 = 5. For n = 2, a₃ = 2a₂ + 2 = 2(5) + 2 = 12. For n = 3, a₄ = 2a₃ + 3 = 2(12) + 3 = 27. Thus the sequence starts 2, 5, 12, 27, and option C is correct. The other choices can result from omitting an index, using the wrong preceding term, or stopping the recurrence too early. The changing added value 1, 2, 3 is essential.

In a recursive rule, each new term is calculated from the preceding term. Thus, a_2=2(5)-2(1)=8, a_3=2(8)-2(2)=12, and a_4=2(12)-2(3)=18. Therefore, 18 is correct. A value such as 20 may result from forgetting to subtract 2n in the final step. Exam tip: to find a_4, substitute n=1, then 2, then 3 in order.

Using the recursive rule with \(n=1\), \(a_2=2(5)-2(1)=8\). Next, with \(n=2\), \(a_3=2(8)-2(2)=12\). Hence, 12 is correct. The answer 14 can result from incorrectly subtracting 2 in the second step; since \(n=2\), the term \(2n\) equals 4. Exam tip: write the value of \(n\) separately at every step of a recursive rule.

The answer is option C, 26. The rule uses the previous term and the term before that: a_n=a_{n-1}+2a_{n-2}. Start with a_1=1 and a_2=4. For n=3, a_3=a_2+2a_1=4+2(1)=6. For n=4, a_4=a_3+2a_2=6+2(4)=14. For n=5, a_5=a_4+2a_3=14+2(6)=26. Option A, 18, is not obtained because the second previous term may have been used incorrectly. Option B, 22, is also not the result of the stated rule. Option C, 26, follows exactly from the two required earlier terms. Option D, 30, is too large and comes from an arithmetic or indexing mistake. The key detail is that 2 multiplies a_{n-2}, not a_{n-1}. Write the two earlier terms beside each new calculation to avoid mixing their positions.

Using the recursive rule, each new term is found by doubling the previous term and subtracting the term before it. Thus, \(a_3=2(5)-2=8\), \(a_4=2(8)-5=11\), \(a_5=2(11)-8=14\), and \(a_6=2(14)-11=17\). Therefore, 17 is correct. The close distractor 14 is \(a_5\), not \(a_6\). Exam tip: when using a two-term recursive rule, write the term number beside each calculated term.

The recursive rule adds the square of the current index to obtain the next term. Thus, \(a_2=3+1^2=4\), \(a_3=4+2^2=8\), \(a_4=8+3^2=17\), and \(a_5=17+4^2=33\). Therefore, 33 is correct. A value such as 30 would result from using an incorrect increment instead of the required squares. Exam tip: to find \(a_5\), add only \(1^2\) through \(4^2\).

In the recursive rule, add the square of the current value of \(n\) at each step. Thus, \(a_2=3+1^2=4\), \(a_3=4+2^2=8\), and \(a_4=8+3^2=17\). Therefore, the correct answer is 17. A value such as 14 can result from adding incorrect values instead of the required squares. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).

Apply the recursive rule successively for n=1, 2, and 3. a_2=50-(2×1+3)=45, a_3=45-(2×2+3)=38, and a_4=38-(2×3+3)=29. Therefore, 29 is correct. The value 31 can result from incorrectly subtracting 7 instead of 9 in the third step. Exam tip: starting from a_1, apply the rule three times to find a_4.

In the recursive rule, each new term is obtained by multiplying the previous term by 3 and adding 1. Thus, a_2=3(2)+1=7, a_3=3(7)+1=22, and a_4=3(22)+1=67. Therefore, 67 is correct. A value such as 64 may result from forgetting to add 1 in the final step. Exam tip: To find a_4 from a_1, apply the rule successively three times.

Given a₁ = 2, apply the recursive rule: a₂ = 3 × 2 + 1 = 7. Then a₃ = 3 × 7 + 1 = 22. Therefore, 22 is correct. A value such as 20 results from an error in multiplying by 3 or adding 1 while finding the next term. Exam tip: for each new term, multiply the previous term by 3 first, then add 1.

Apply the recursive rule step by step, adding the current value of \(n\) after halving the preceding term. \(a_2=\frac{10}{2}+1=6\), \(a_3=\frac{6}{2}+2=5\), and \(a_4=\frac{5}{2}+3=5.5\). Hence, 5.5 is correct. The value 6.5 would result from incorrectly adding \(4\) in the last step; to find \(a_4\), the rule uses \(n=3\). Exam tip: in \(a_{n+1}\), use an \(n\) value one less than the index of the term being found.

Use \(n=1\) to find the second term: \(a_2=\frac{10}{2}+1=6\). Then use \(n=2\) to find the third term: \(a_3=\frac{6}{2}+2=5\). Therefore, the correct answer is 5. Note that 6 is \(a_2\), not \(a_3\). Exam tip: when finding \(a_{n+1}\), use an \(n\)-value one less than the index of the term required.

Using the recursive rule, \(a_3=2(3)+1=7\), \(a_4=2(7)+3=17\), and \(a_5=2(17)+7=41\). Hence, the correct answer is 41. The close distractor 17 is \(a_4\), not \(a_5\). Exam tip: write each term in sequence and check the subscript carefully.

First find the third term using the recursive rule: \(a_3=2a_2+a_1=2(3)+1=7\). Then \(a_4=2a_3+a_2=2(7)+3=17\). Therefore, 17 is correct. A value such as 15 can result from not multiplying the previous term correctly by 2. Exam tip: write every intermediate term step by step until the required term is reached.

Substituting n=1,2,3,4 in the recursive rule gives successive additions of 2, 5, 8, and 11. Hence the terms are 4, 6, 11, 19, 30, so a_5=30. The value 32 would require using 13 instead of 11 in the final step, which does not follow the rule. Exam tip: use the next value of n at each recursive step.

In a recursive rule, use the next value of n at each step. For n=1, a_2=4+3(1)-1=6; for n=2, a_3=6+3(2)-1=11; and for n=3, a_4=11+3(3)-1=19. Therefore, 19 is correct. A choice such as 17 may result from not updating the value of 3n-1 at every step. Exam tip: To find a_4, apply the rule successively for n=1, 2, and 3.

In a recursive rule, each new term is calculated from the previous term. First, \(a_2=a_1^2-1=2^2-1=3\). Then \(a_3=a_2^2-1=3^2-1=8\). Therefore, the correct answer is 8. An answer such as 6 may result from subtracting 1 without correctly squaring the previous term. Exam tip: always find \(a_2\) before calculating \(a_3\).