Recursive rule

The recursive rule adds the previous two terms and 7. Thus, \(a_3=3+9+7=19\), \(a_4=9+19+7=35\), and \(a_5=19+35+7=61\). Therefore, the correct answer is 61. The option 54 is only \(19+35\), so it misses the required \(+7\). Exam tip: at every step, include both previous terms and the constant 7.

In a recursive rule, each new term is found from the preceding term. First, \(a_2=7\times2-4=10\). Then, \(a_3=7\times10-4=66\). Therefore, the correct answer is 66. A value such as 62 can result from an incorrect multiplication or subtraction in the final step. Exam tip: always write \(a_2\) before finding \(a_3\).

The recursive rule adds \(9n-4\) to the current term at each step. For \(n=1\), \(a_2=6+5=11\); for \(n=2\), \(a_3=11+14=25\); and for \(n=3\), \(a_4=25+23=48\). Hence, 48 is correct. A choice such as 45 may result from not using the successive values of \(n\) correctly. Exam tip: to find \(a_4\), apply the rule successively with \(n=1,2,3\).

Apply the recursive rule successively for \(n=1,2,3\). The amounts subtracted are \(13,20,27\), so \(a_2=210-13=197\), \(a_3=197-20=177\), and \(a_4=177-27=150\). Hence, option A is correct. A nearby value such as \(153\) can result from calculating \(7n+6\) incorrectly in one step. Exam tip: to find \(a_4\) from \(a_1\), apply the rule three times.

First find the third term using the recursive rule: \(a_3=5(3)-4(2)+3=10\). Then, \(a_4=5(10)-4(3)+3=50-12+3=41\). Therefore, 41 is correct. The nearby distractor 44 can result from not subtracting the \(4a_{n-2}\) term correctly. Exam tip: substitute both previous terms carefully before simplifying.

The total addition is (\frac{7(1+2+3+4)}{2}=35), so (a_5=42). Exam tip: find total fractional addition first.

The governing concept is recursive evaluation with an index-dependent fractional decrement. To move from a₁ to a₅, apply the rule for n = 1, 2, 3, and 4. The total amount subtracted is (7/2)(1 + 2 + 3 + 4) = (7/2)(10) = 35. Therefore a₅ = 80 − 35 = 45, so option A is correct. Checking each step gives a₂ = 80 − 7/2 = 76.5, a₃ = 76.5 − 7 = 69.5, a₄ = 69.5 − 10.5 = 59, and a₅ = 59 − 14 = 45. Skipping a transition or mishandling the fraction produces the distractors.

In the recursive rule, use the current index n at each step. Thus, \(a_2=1\times4+5=9\), \(a_3=2\times9+5=23\), and \(a_4=3\times23+5=74\). Therefore, 74 is correct. An answer such as 71 results from an error in the multiplication or addition in the final step. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).

Apply the recursive rule successively using the appropriate value of n at each step. For n=1, a_2=3×5−1=14; for n=2, a_3=4×14−1=55; and for n=3, a_4=5×55−1=274. Hence, the correct answer is 274. A value such as 280 results from forgetting to subtract 1 in the last step. Exam tip: to find a_4, apply the rule in order for n=1, 2, and 3.

Each term is obtained by multiplying the preceding term by 3 and then adding 2: \(3\times4+2=14\), \(3\times14+2=44\), and \(3\times44+2=134\). Also, the first term is \(a_1=4\). Therefore, the correct recursive rule is \(a_1=4,\ a_{n+1}=3a_n+2\). Under option D, the next term would be \(3\times4-2=10\), not 14. Exam tip: test a recursive rule using at least two consecutive terms.

A recursive rule must reproduce the first term and every successive difference. The sequence differences are 11 − 6 = 5, 19 − 11 = 8, and 30 − 19 = 11. These increments follow 3n + 2 for n = 1, 2, and 3. Starting with a₁ = 6 gives the required sequence, so option D is correct. Options A and C give different increments, while B begins with the wrong first term.

The governing concept is a recursive sequence whose increment depends on the index n. Start with a₁ = 5 and apply the rule aₙ₊₁ = aₙ + n² one step at a time. For n = 1, a₂ = 5 + 1² = 6. For n = 2, a₃ = 6 + 2² = 10. For n = 3, a₄ = 10 + 3² = 19. For n = 4, a₅ = 19 + 4² = 35. For n = 5, a₆ = 35 + 5² = 60. Therefore 60 is the sixth term, so option A is correct. The increments are 1, 4, 9, 16, and 25, the consecutive squares required by the recurrence. The seventh, eighth, and ninth terms come after a₆ and are larger than 60, so options B, C, and D cannot be the first position of 60.

The rule adds two quantities at every step: n! and 2ⁿ. To obtain a₅ from a₁, apply the rule for n = 1, 2, 3, and 4. Both parts must be calculated separately and then added. The supplied explanation lists the second added value incorrectly: for n = 2, 2! + 2² = 2 + 4 = 6 is actually correct, but the resulting sequence must still be checked carefully.

The additions are 3, 6, 14, and 40. Thus a₂ = 5 + 3 = 8, a₃ = 8 + 6 = 14, a₄ = 14 + 14 = 28, and a₅ = 28 + 40 = 68. Equivalently, 5 + 3 + 6 + 14 + 40 = 68. Therefore option B is correct, and the supplied answer is consistent. The key is to distinguish the added amounts from the sequence terms; the listed additions are not themselves the full sequence.

The recurrence tells us how to obtain the next term from the current term. To find \\(a_5\\), begin with \\(a_1=120\\) and apply the rule for \\(n=1,2,3,4\\). In every step, the same index \\(n\\) must be used in both \\(n!\\) and \\(2^n\\). The subtraction is applied four times because four transitions lead from the first term to the fifth.

For \\(n=1\\), subtract \\(1!+2^1=1+2=3\\), so \\(a_2=117\\). For \\(n=2\\), subtract \\(2!+2^2=2+4=6\\), giving \\(a_3=111\\). For \\(n=3\\), subtract \\(6+8=14\\), giving \\(a_4=97\\). For \\(n=4\\), subtract \\(24+16=40\\), so \\(a_5=57\\). Hence option D is correct.

Substituting \(n=1,2,3\) in the recursive rule gives \(a_2=2(4)-3=5\), \(a_3=2(5)+5=15\), and \(a_4=2(15)-7=23\). Therefore, the correct answer is \(23\). A nearby choice such as \(25\) can result from missing that \((-1)^3=-1\), so \(7\) must be subtracted. Exam tip: determine the sign of \((-1)^n\) before substituting each value of \(n\).

Apply the recursive rule for the first three steps. For n=1, (a_2=50-(-1)^1(1^2+1)=52). For n=2, (a_3=52-(+1)(2^2+1)=47). For n=3, (a_4=47-(-1)(3^2+1)=57). Therefore, the correct answer is 57. A value such as 60 can result from mishandling the outer minus sign together with (-1)^n. Exam tip: substitute n step by step and determine the sign of (-1)^n before simplifying.

Using the recursive rule successively for n=3,4,5: a_3=2(5)+2+6=18, a_4=2(18)+5+8=49, and a_5=2(49)+18+10=126. Therefore, the correct answer is 126. A value such as 118 can result from adding the 2n term incorrectly. Exam tip: use the current value of n when calculating 2n at every step.

The recurrence adds the fixed first term \(a_1=6\) at every step. For \(n=1\), \(a_2=6+6+1(2)=14\); for \(n=2\), \(a_3=14+6+2(3)=26\); and for \(n=3\), \(a_4=26+6+3(4)=44\). Therefore, the correct answer is 44. A value such as 47 can result from incorrectly evaluating \(n(n+1)\) or using the wrong step number. Exam tip: to find \(a_4\), apply the rule successively with \(n=1,2,3\).

Substitute \(n=1,2,3\) in the recursive rule. The successive increments are \(3^2=9\), \(5^2=25\), and \(7^2=49\). Thus, \(a_2=3+9=12\), \(a_3=12+25=37\), and \(a_4=37+49=86\). Hence, 86 is correct. The value 83 would result from using an incorrect final increment. Exam tip: to find \(a_4\) from \(a_1\), apply the rule three times.

Here each step subtracts the square of an odd number of the form 2n + 1. To reach a₄ from a₁, use n = 1, 2, and 3. The corresponding odd numbers are 3, 5, and 7, so the subtractions are 3² = 9, 5² = 25, and 7² = 49. The changing value of n must be used at each step.

Starting from 180, subtract 9 to get a₂ = 171. Subtract 25 to get a₃ = 146. Finally, subtract 49 to obtain a₄ = 97. The total subtraction is 9 + 25 + 49 = 83, and 180 − 83 = 97 as well. Therefore option D is correct. The values 9, 25, and 49 are squares, not the odd numbers themselves; confusing those two levels would produce an incorrect result.