Recursive rule

Using the recursive rule, \(a_3=2(5)-1+2=11\), \(a_4=2(11)-5+2=19\), and \(a_5=2(19)-11+2=29\). Therefore, 29 is the correct answer. A value such as 26 can result from subtracting the wrong preceding term. Exam tip: write \(a_{n-1}\) and \(a_{n-2}\) separately, then multiply, subtract, and finally add 2.

The recurrence adds a changing amount to the current term. To find \(a_5\), four increments are needed: use \(n=1,2,3,4\) in turn. Simplifying the fraction before adding makes the work clear: the increments are \(\frac{1(4)}2=2\), \(\frac{2(5)}2=5\), \(\frac{3(6)}2=9\), and \(\frac{4(7)}2=14\). The recurrence is used four times because the starting term is \(a_1\).

Now calculate \(a_2=4+2=6\), \(a_3=6+5=11\), \(a_4=11+9=20\), and \(a_5=20+14=34\). Equivalently, add the increments: \(4+2+5+9+14=34\). Therefore option C is correct. The listed increments are correct, and careful indexing shows why there are four of them.

Apply the recursive rule successively for \(n=1,2,3\). \(a_2=90-\frac{1(1+3)}{2}=88\), \(a_3=88-\frac{2(2+3)}{2}=83\), and \(a_4=83-\frac{3(3+3)}{2}=83-9=74\). Hence, the correct answer is \(74\). A value such as \(76\) can result from calculating a subtracted term incorrectly. Exam tip: to find \(a_4\), apply the rule three times starting from \(a_1\).

Use the recursive rule first with \(n=1\): \(a_2=6+\frac{6}{2}+1=10\). Then use \(n=2\): \(a_3=10+\frac{10}{2}+2=17\). Hence, 17 is correct. The value 15 would result from omitting the \(n\) term, so it is not correct. Exam tip: while finding successive terms from \(a_{n+1}\), update the value of \(n\) as \(1,2,\ldots\).

The recurrence contains the current term itself, so each step adds one-fourth of that current term and also adds n. It is important not to use a₁ again in the second step; the rule must use the term just obtained. To reach a₃ from a₁, calculate exactly two successive terms.

With a₁ = 4 and n = 1, a₂ = 4 + 4/4 + 1 = 4 + 1 + 1 = 6. Next use n = 2 and the current term a₂ = 6: a₃ = 6 + 6/4 + 2 = 6 + 3/2 + 2 = 19/2. Thus the correct value is 19/2, which is option B. Writing 6/4 as 3/2 avoids a decimal approximation and keeps the result exact. The other listed fractions do not follow from the recurrence.

Given \(a_1=2\), first use \(n=1\) in the recursive rule: \(a_2=a_1^2+1=2^2+1=5\). Next, with \(n=2\), \(a_3=a_2^2+2=5^2+2=27\). Therefore, 27 is correct. The value 25 is only \(5^2\); the required addition of 2 has been omitted. Exam tip: to find \(a_3\), calculate \(a_2\) first and use the correct value of \(n\) at each step.

Using the recursive rule with n=1, a₂=a₁²−1=3²−1=8. Then, with n=2, a₃=a₂²−2=8²−2=64−2=62. Therefore, 62 is the correct answer. 64 is a close distractor because it results from forgetting to subtract 2. Exam tip: while finding the next term, use n equal to one less than that term’s index.

In C, \(a_n\) uses both \(a_{n-1}\) and \(n\); for example, \(a_3=a_2+3\). A and B use only the preceding term. Exam tip: check whether \(n\) appears in the recurrence.

(a_3=3\times4-3+3=12) and (a_4=3\times12-4+4=36). Exam tip: use both previous terms and current (n).

Apply the recursive rule successively for n=1, 2, and 3: a_2=2(5)+(-1)^1=9, a_3=2(9)+(-1)^2=19, and a_4=2(19)+(-1)^3=37. Therefore, 37 is correct. The nearby option 39 results from incorrectly using +1 instead of (-1)^3=-1 in the last step. Exam tip: substitute the value of n at every step to check the sign of (-1)^n.

Substitute n=1,2,3 in the recursive rule. We get a_2=30-(-1)^1(3)=33, a_3=33-(-1)^2(5)=28, and a_4=28-(-1)^3(7)=35. Therefore, the correct answer is 35. The value 33 is only the second term, not the fourth term. Exam tip: determine the sign of (-1)^n first, then apply the outer minus sign.

The added values (1!,2!,3!,4!) are (1,2,6,24), so (a_5=36). Exam tip: add factorial values in order.

A factorial means multiplying all positive integers up to the given number: 1! = 1, 2! = 2, 3! = 6, and 4! = 24. Because the requested term is a₅ and the initial term is a₁, the rule aₙ₊₁ = aₙ − n! is applied for n = 1, 2, 3, and 4. Every factorial is subtracted from the term already obtained.

Starting with 100, subtract 1! to get a₂ = 99. Subtract 2! = 2 to get a₃ = 97. Then subtract 3! = 6 to get a₄ = 91. Finally subtract 4! = 24 to obtain a₅ = 67. The same result comes from 100 − (1 + 2 + 6 + 24) = 100 − 33 = 67. Therefore option C is correct. Listing factorials in order is a reliable way to avoid skipping a step.

Use the value of n for each step in the recursive rule. Starting with a_1=1, for n=1, a_2=2(1)+1^2=3. Then, for n=2, a_3=2(3)+2^2=10, and for n=3, a_4=2(10)+3^2=29. Therefore, 29 is correct. A value such as 31 can result from using an incorrect square in the final step. Exam tip: increase n one step at a time while finding a_{n+1}.

Here, \(a_{n+1}-a_n=2n\). At the next step the difference is \(2(n+1)\), which is 2 greater. Thus the differences are not constant, so it is not an arithmetic progression. Exam tip: write consecutive differences first.

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.

Direct answer: C, 15. Notice that the formula contains two different symbols: aₙ is the current term, while a₁ is the fixed first term. Since a₁=8, it remains 8 in every step. First, 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. Hence C is correct. A, 13, does not result from the stated divisions and additions. B, 14, is the value of a₃, so it stops one step too early. C, 15, correctly uses a₃ as the current term and the fixed value a₁=8. D, 16, can arise if the fixed first term is used incorrectly or if the final division is mishandled. The sequence is 8, 12, 14, 15. Common warning: a₁ is not replaced by a₂ or a₃; it always denotes the original first term.

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 direct answer is option D, \(a_4=66\). The rule is \(a_{n+1}=a_n-(2^n+n^2+n)\), so for each step calculate the entire expression in parentheses first, then subtract it from the current term. Starting with \(a_1=100\), for \(n=1\), the amount subtracted is \(2^1+1^2+1=2+1+1=4\), giving \(a_2=100-4=96\). For \(n=2\), subtract \(2^2+2^2+2=4+4+2=10\), giving \(a_3=96-10=86\). For \(n=3\), subtract \(2^3+3^2+3=8+9+3=20\), giving \(a_4=86-20=66\). Option A, 58, subtracts too much. Option B, 62, does not follow the three correct steps. Option C, 64, also reflects an arithmetic error. Option D, 66, is correct. A useful check is that the successive subtractions are 4, 10, and 20, whose total is 34; hence \(100-34=66\).

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.