Recursive rule

The recursive rule uses the current index n at each step. Thus, a_2=1×3+2=5, then a_3=2×5+4=14, and a_4=3×14+6=48. Therefore, the correct answer is 48. A value such as 45 may result from using an incorrect index in the final step. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

In the recursive rule, the value of n changes at every step. For n=1, a_2=2a_1-1=2×2-1=3. Then, for n=2, a_3=3a_2-1=3×3-1=8, and for n=3, a_4=4a_3-1=4×8-1=31. Therefore, the correct answer is 31. The value 33 would result from forgetting to subtract 1 in the final step. Exam tip: write the current index n before applying a recursive rule to check the multiplier.

To obtain a₄, apply the rule for n = 1, 2, and 3. The corresponding subtractions are 2¹ + 1² = 3, 2² + 2² = 8, and 2³ + 3² = 17. Thus a₂ = 200−3 = 197, a₃ = 197−8 = 189, and a₄ = 189−17 = 172. Option C is correct. Confusing powers with products or omitting the square term leads to the distractors.

Apply the recursive rule successively for n=1, 2, and 3: \(a_2=1+3^1+1=5\), \(a_3=5+3^2+2=16\), and \(a_4=16+3^3+3=46\). Hence, 46 is correct. The value 49 would result from incorrectly adding an extra amount in the final step. Exam tip: to find \(a_4\), use the recurrence consecutively from n=1 through n=3.

(a_3=2\times3+3\times2=12) and (a_4=2\times12+3\times3=33). Exam tip: apply coefficients to the correct terms.

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.