Recursive rule

Using the recursive rule successively for n=1, 2, and 3: a_2=6+5(1)-2=9, a_3=9+5(2)-2=17, and a_4=17+5(3)-2=30. Therefore, the correct answer is 30. A value such as 29 can result from incorrectly evaluating 5n-2 in the final step. Exam tip: to find a_4, apply the rule step by step from n=1 to n=3.

Use the recursive rule one step at a time. For \(n=1\), \(a_2=6+5(1)-2=9\). Then, for \(n=2\), \(a_3=9+5(2)-2=17\). Hence, 17 is correct. The value 19 can result from using an incorrect value of \(n\) in the second step. Exam tip: substitute the correct value of \(n\) at each stage of a recursive rule.

Apply the recursive rule successively for n=1, 2, and 3. We get a_2=1+(1^2+1)=3, a_3=3+(2^2+2)=9, and a_4=9+(3^2+3)=21. Therefore, the correct answer is 21. The value 23 would result from incorrectly using 14 instead of 12 as the final increment. Exam tip: To find a_4, apply the rule only from n=1 through n=3.

Using n=1 first, a_2=a_1+1^2+1=1+2=3. Then, using n=2, a_3=a_2+2^2+2=3+6=9. Therefore, the correct answer is 9. The value 7 can result from using an incorrect increment instead of 2^2+2. Exam tip: substitute the next value of n separately for each new term.

The recursive rule is \(a_{n+1}=a_n+2a_n=3a_n\). Hence, \(a_2=3\times9=27\) and \(a_3=3\times27=81\). Therefore, \(81\) is correct. An option such as \(72\) does not apply the rule of multiplying each term by 3 correctly. Exam tip: first combine like terms to simplify a recursive rule.

This is a recursive rule because each new term is calculated from the immediately preceding term. Begin with a₁ = 12. For n = 1, a₂ = a₁ + a₁/2 = 12 + 6 = 18. The next step must use a₂, not the original a₁: a₃ = a₂ + a₂/2 = 18 + 9 = 27. Thus option B is correct. An equivalent view is that each term is multiplied by 1 + 1/2 = 3/2. Consequently, a₂ = 12 × 3/2 = 18 and a₃ = 18 × 3/2 = 27. Option A is only the result after the first update, so it is a₂. Options C and D do not follow from adding half of the current term and therefore violate the given recurrence.

In the recursive rule, add \(3n\) to the current term at each step. \(a_2=4+3(1)=7\), \(a_3=7+3(2)=13\), and \(a_4=13+3(3)=22\). Therefore, the correct answer is 22. A value such as 21 can result from incorrectly treating the increase as constant. Exam tip: to find \(a_4\), substitute \(n=1,2,3\) in order.

Using n=1, 2, and 3 in the recursive rule, the quantities subtracted are 3, 5, and 7 respectively. Thus, a_2=30-3=27, a_3=27-5=22, and a_4=22-7=15. Therefore, the correct answer is 15. The value 17 would result from subtracting 5 in the final step, but for a_4, n=3, so 7 must be subtracted. Exam tip: write the value of n for every step before applying the rule.

Use the recursive rule successively, remembering that n changes at every step. From a₁ = 2, take n = 1: a₂ = 2a₁ + 1 = 2(2) + 1 = 5. Next, with n = 2, a₃ = 2a₂ + 2 = 2(5) + 2 = 12. With n = 3, a₄ = 2a₃ + 3 = 2(12) + 3 = 27. Finally, use n = 4 to obtain a₅ = 2a₄ + 4 = 2(27) + 4 = 54 + 4 = 58. Hence option D is correct, and the sequence is 2, 5, 12, 27, 58. The alternatives 48, 52, and 56 may arise from stopping early, adding the wrong index, or making an intermediate arithmetic error. The recurrence must be applied four times to reach the fifth term.

In a recursive rule, each new term is calculated from the preceding term. First, \(a_2=3\times6-2=16\). Then, \(a_3=3\times16-2=46\), so 46 is correct. A result of 48 can occur if the subtraction is omitted. Exam tip: to find \(a_3\), apply the rule first to \(a_1\) and then to \(a_2\).

In the recursive rule, each new term is obtained by adding n²+1 to the preceding term for the current value of n. Thus, a₂=1+1²+1=3, a₃=3+2²+1=8, and a₄=8+3²+1=18. Therefore, 18 is correct. An answer such as 16 can result from using an incorrect value of n or an incorrect increment. Exam tip: to find a₄, apply the rule successively for n=1, 2, and 3.

The recursive rule subtracts \(n^2\) to obtain each next term. Thus, \(a_2=50-1^2=49\), \(a_3=49-2^2=45\), \(a_4=45-3^2=36\), and \(a_5=36-4^2=20\). Therefore, the correct answer is \(20\). The nearby choice \(21\) could result from incorrectly subtracting \(15\) instead of \(4^2=16\). Exam tip: to find \(a_5\), apply the rule successively for \(n=1,2,3,4\).

Using the recursive rule, \(a_3=5+2(3)=11\), \(a_4=11+2(5)=21\), and \(a_5=21+2(11)=43\). Therefore, the correct answer is 43. An option such as 45 may result from applying the rule incorrectly to the preceding terms. In exams, check the coefficient of each previous term before calculating the next term.

This is a two-term recursive sequence: every term from the third onward is obtained by adding the previous term to twice the term before it. The order of the terms matters, so the formula must be applied with the correct indices. We are given \(a_1=3\) and \(a_2=5\).

Using the rule, \(a_3=a_2+2a_1=5+2(3)=11\). Next, \(a_4=a_3+2a_2=11+2(5)=21\). Therefore the fourth term is 21. A common error is to use \(a_1\) again when finding \(a_4\), but the rule requires the two immediately preceding terms, \(a_3\) and \(a_2\). Hence option C is correct.

Using the recurrence, \(a_3=2(9)-4=14\), then \(a_4=19\), \(a_5=24\), and \(a_6=29\). Therefore, 29 is correct. The consecutive terms increase by 5; 24 is a close distractor, but it is \(a_5\), not \(a_6\). Exam tip: check the difference between the first two terms to identify the pattern quickly.

The recursive rule adds successive powers of 2 to the preceding term. Thus, \(a_2=2+2^1=4\), \(a_3=4+2^2=8\), \(a_4=8+2^3=16\), and \(a_5=16+2^4=32\). Therefore, 32 is correct. Although 34 is close, it does not result from adding \(2^4=16\) to the previous term. Exam tip: to find \(a_5\), apply the rule successively from \(n=1\) to \(n=4\).

The amount subtracted changes at each step. \(a_2=90-2^1=88\), \(a_3=88-2^2=84\), and \(a_4=84-2^3=76\). Therefore, the correct answer is \(76\). The value \(78\) could result from incorrectly subtracting 2, 4, and 6 instead of powers of 2. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).

Substituting successive values of n in the recursive rule gives a₂=5+1=6, a₃=6−2=4, a₄=4+3=7, and a₅=7−4=3. Therefore, the correct answer is 3. Option 4 is a close distractor because it is the value of a₃, not a₅. Exam tip: check whether the exponent of (-1) is even or odd before deciding the sign.

Apply the recursive rule successively for \(n=1,2,3\). \(a_2=5+(-1)^2\cdot1=6\), \(a_3=6+(-1)^3\cdot2=4\), and \(a_4=4+(-1)^4\cdot3=7\). Therefore, the correct answer is \(7\). Option \(6\) is only the second term, not the fourth term. Exam tip: check the sign of \((-1)^{n+1}\) separately at every step.

Apply the recursive rule successively for n=1, 2, and 3. We get a_2=2+1(1+2)=5, a_3=5+2(2+2)=13, and a_4=13+3(3+2)=28. Therefore, the correct answer is 28. The value 26 is not obtained because the increment in the third step is 3(5)=15. Exam tip: To find a_4 starting from a_1, apply the rule three times.

Apply the recursive rule first with n=1: a_2=2+1(1+2)=5. Then use n=2: a_3=5+2(2+2)=5+8=13. Therefore, 13 is correct. A value such as 11 can result from calculating the second increment incorrectly. Exam tip: to find a_3, apply the rule successively for n=1 and n=2.

Using n=1, 2, and 3 in the recursive rule, the values subtracted are 1(3)=3, 2(4)=8, and 3(5)=15. Thus, a_2=100-3=97, a_3=97-8=89, and a_4=89-15=74. Therefore, 74 is correct. A value such as 76 would result from an incorrect final subtraction. Exam tip: to find a_4 from a_1, apply the rule three times.

Apply the recursive rule step by step: \(a_2=2(7)+3=17\), \(a_3=2(17)+3=37\), and \(a_4=2(37)+3=77\). Therefore, the correct answer is 77. An option such as 75 can result from an arithmetic error in the final step. In exams, write each previous term before calculating the next one.

Each new term is obtained by doubling the previous term and adding 3. Thus, \(a_2=2\times7+3=17\), and \(a_3=2\times17+3=37\). Therefore, 37 is correct. The value 35 would result from an incorrect calculation in the second step. Exam tip: in a recursive sequence, find the terms one step at a time rather than jumping directly to a later term.

In A, \(a_{n+1}-a_n=5\) for every \(n\), so the sequence is an AP. In B, the difference is \(a_n+5\), which is not fixed. Exam tip: check the difference of consecutive terms.