Recursive rule

In the correct rule, each new term is obtained by multiplying the previous term by 3 and adding 1: \(3\times4+1=13\), \(3\times13+1=40\), and \(3\times40+1=121\). Hence, \(a_1=4,\ a_{n+1}=3a_n+1\) is correct. Option B subtracts 1 and gives 11 after 4. Option D has the same recurrence but an incorrect first term, 13. Exam tip: verify both the initial term and at least two successive transitions in a recursive sequence.

The correct rule starts with \(a_1=9\). For \(n=1\), \((-1)^1(2)=-2\), so \(9-2=7\). For \(n=2\), adding \(+3\) gives 10, and for \(n=3\), adding \(-4\) gives 6. Hence, option A is correct. Option D gives the first change of \(-2\), but its next change would be \(-4\), whereas the sequence next changes by \(+3\). Exam tip: test a recursive rule by substituting \(n=1,2,3\) to generate the initial terms.

The governing idea is repeated substitution in a recurrence, with a fixed initial term appearing in the formula. Since a₁ = 8, the second occurrence of a₁ remains 8 at every step; it must not be replaced by the current term. Thus a₂ = a₁/2 + a₁ = 8/2 + 8 = 12. Next, a₃ = a₂/2 + a₁ = 12/2 + 8 = 14. Finally, a₄ = a₃/2 + a₁ = 14/2 + 8 = 15. Therefore option C is correct. The generated sequence is 8, 12, 14, 15. Replacing the fixed a₁ by aₙ, or stopping after a₃, leads to an incorrect answer.

In the recursive rule, add \(na_1\) to the previous term, while \(a_1=7\) remains fixed. Thus, \(a_2=7+1\times7=14\), \(a_3=14+2\times7=28\), and \(a_4=28+3\times7=49\). Therefore, 49 is the correct answer. One may incorrectly get 56 by adding \(4\times7\) in the last step, but for finding \(a_4\), the rule uses \(n=3\). Exam tip: in \(a_{n+1}\), substitute a value of \(n\) that is one less than the required subscript.

This is a two-term recursive relation with an additional index term n. The initial values are a₁ = 2 and a₂ = 5. To find a₃, put n = 3 into the recurrence: a₃ = 2a₂ + a₁ + 3 = 2(5) + 2 + 3 = 15. Next, put n = 4: a₄ = 2a₃ + a₂ + 4 = 2(15) + 5 + 4 = 39. Hence option C is correct. The term n must be added after using the two preceding terms. Omitting it gives 35, while other distractors arise from using the wrong index or an incorrect preceding term. The calculated terms 2, 5, 15, 39 confirm the answer.

The subtracted values are (4,10,20), so (a_4=100-34=66). Exam tip: add the power, square, and (n) before subtracting.

In the recursive rule, double the current term and then add the square of the step number. \(a_2=2(5)+1^2=11\), \(a_3=2(11)+2^2=26\), and \(a_4=2(26)+3^2=61\). Therefore, 61 is correct. A value such as 59 results from not adding \(3^2\) correctly in the final step. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).

The subtracted values are (3,10,21), so (a_4=66). Exam tip: add all subtractions first and then subtract.

For \(n=1\), \(a_2=100-(2\times1^2+1)=97\). Then, for \(n=2\), \(a_3=97-(2\times2^2+2)=97-10=87\). Therefore, the correct answer is 87. The value 89 would result from calculating the amount subtracted in the second step incorrectly. Exam tip: in a recursive rule, substitute the new value of \(n\) for each successive term.

In this recursive rule, each new term is formed using the two immediately preceding terms. Thus, \(a_3=2\times7+2=16\), \(a_4=2\times16+7=39\), and \(a_5=2\times39+16=94\). Therefore, 94 is correct. A value such as 96 can result from an arithmetic error; both \(a_4\) and \(a_3\) must be used to find \(a_5\). Exam tip: write each term in order before applying the recurrence again.

A recursive rule finds each new term from earlier terms. Here the rule is \\(a_n=2a_{n-1}+a_{n-2}\\), so twice the immediately preceding term is added to the term before it. The coefficient 2 does not multiply both terms. Starting with a₁ = 2 and a₂ = 7, we can calculate the terms in order.

First, \\(a_3=2a_2+a_1=2(7)+2=16\\). Next, \\(a_4=2a_3+a_2=2(16)+7=39\\). Thus the fourth term is 39, which is option C. A common error is to use only the previous term or to multiply the sum of the two previous terms by 2; neither follows the stated recurrence.

Because this is a second-order recurrence, calculate each term from the two immediately preceding terms. First, a₃ = 3(10) − 2(4) = 30 − 8 = 22. Next, a₄ = 3(22) − 2(10) = 66 − 20 = 46. Finally, a₅ = 3(46) − 2(22) = 138 − 44 = 94. Therefore option C is correct; applying both coefficients to the wrong terms changes the result.

The added values are (4,13,36), so (a_4=56). Exam tip: calculate the power and square separately.

Apply the recursive rule step by step. \(a_2=3+3^1+1^2=7\), and \(a_3=7+3^2+2^2=7+9+4=20\). Hence, 20 is correct. A value such as 18 can result from omitting \(2^2\), which is incorrect. Exam tip: use the current value of \(n\) in \(a_{n+1}\); for \(a_3\), take \(n=2\).

The subtracted values are (3,8,17,32), so (a_5=90). Exam tip: add all subtractions before subtracting.

Apply the recurrence for n = 1, 2, and 3. The deductions are 2¹ + 1² = 3, 2² + 2² = 8, and 2³ + 3² = 17. Starting from 150 gives a₂ = 150−3 = 147, a₃ = 147−8 = 139, and a₄ = 139−17 = 122. Hence option B is correct. The other values arise from an arithmetic error or from using the wrong index in the power or square.

Substituting n=1,2,3,4 in the recursive rule gives: a_2=12-4=8, a_3=8+7=15, a_4=15-10=5, and a_5=5+13=18. Hence, the correct answer is 18. The value 15 is a close distractor because it is a_3, not a_5. Exam tip: track the alternating sign from (-1)^n carefully: it is negative for odd n and positive for even n.

Apply the recursive rule successively for n=1, 2, and 3. We get a_2=12-4=8, a_3=8+7=15, and a_4=15-10=5. Therefore, the correct answer is 5. The value 7 may result from not applying the third-step change correctly. Exam tip: determine the sign of (-1)^n first at every step, then calculate the next term.

The added values are (7,14,23), so (a_4=50). Exam tip: calculate the product and then add (2).

Use the recursive rule successively to find the next terms. For \(n=1\), \(a_2=6+1(1+4)+2=13\). Then, for \(n=2\), \(a_3=13+2(2+4)+2=13+12+2=27\). Therefore, the correct answer is \(27\). The nearby distractor \(25\) can result from incorrectly evaluating \(2(2+4)\) in the second step. Exam tip: substitute the next integer value of \(n\) each time you calculate a new term.