Recursive rule

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 terms are (1,4,6,14,26), so (a_5=26). Exam tip: multiply the second previous term by (2).

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 \(a_3\).

The recursive rule adds a different value at each step. For \(n=1,2,3,4\), the added values are \(3,4,5,6\), respectively. Thus \(a_2=9\), \(a_3=13\), \(a_4=18\), and \(a_5=24\). Hence, 24 is correct; 18 is actually \(a_4\). Exam tip: while finding \(a_{n+1}\), substitute the index \(n\) of the current term first.

In a recursive rule, each new term is found from the preceding term. For n=1, a_2=7+4(1)+1=12; for n=2, a_3=12+4(2)+1=21; and for n=3, a_4=21+4(3)+1=34. Therefore, the correct answer is 34. An answer such as 32 can result from calculating the increment 4n+1 incorrectly. Exam tip: to find a_4, substitute n=1, 2, and 3 successively.

Use the recursive rule first with \(n=1\): \(a_2=\frac{80}{2}-1=39\). Next, use \(n=2\): \(a_3=\frac{39}{2}-2=19.5-2=17.5\). Hence, 17.5 is the correct answer. The option 17 may result from incorrectly treating \(\frac{39}{2}\) as an integer. Exam tip: substitute the correct term number for \(n\) at every step.

The governing idea is to apply the recurrence one step at a time and use the correct index at each step. Starting with a₁ = 84, set n = 1 to calculate a₂: a₂ = a₁/2 − 1 = 84/2 − 1 = 42 − 1 = 41. To calculate a₃, set n = 2, because the rule produces aₙ₊₁ from aₙ: a₃ = a₂/2 − 2 = 41/2 − 2 = 20.5 − 2 = 18.5. Therefore option B is correct. The values 17, 19, and 20 do not follow the stated recurrence. In particular, using n = 1 again or rounding 20.5 before subtracting would produce an incorrect result. The corrected decimal option makes the MCQ mathematically complete.

To find \(a_2\), substitute \(n=1\) in the recursive rule, since \(a_1\) produces the next term \(a_2\). Thus, \(a_2=\frac{a_1}{2}-1=\frac{88}{2}-1=44-1=43\). Option 44 only halves 88; the rule also requires subtracting 1. Exam tip: when finding the next term from \(a_n\), use the index of the current term for \(n\).