Recursive rule

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.

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.

Using the recursive rule successively for \(n=1,2,3,4\), we get \(a_2=3+1^3=4\), \(a_3=4+2^3=12\), \(a_4=12+3^3=39\), and \(a_5=39+4^3=103\). Hence, the correct answer is \(103\). The value \(106\) would result from using an incorrect final increment instead of \(4^3\). Exam tip: to find \(a_5\), add only the increments from \(1^3\) through \(4^3\).

Using n=1, 2, 3 and 4, the terms decrease by 1, 8, 27 and 64 respectively. Thus, a_5=150-(1+8+27+64)=150-100=50. Option 53 results from subtracting only the first three cubes; 4^3 must also be subtracted. Exam tip: to find a_5 from a_1, apply the recursive rule four times.

Apply the recursive rule first for a₃: a₃ = a₂ + 3a₁ + 3 = 6 + 3×4 + 3 = 21. Then, a₄ = a₃ + 3a₂ + 4 = 21 + 3×6 + 4 = 43. Therefore, 43 is correct. A value such as 46 can result from using an incorrect previous term or index value. Exam tip: add the current value of n at every step.

In a recursive rule, substitute the correct value of n at each step. For n=1, a_2=2a_1+1=2×2+1=5. Then, for n=2, a_3=3a_2+2=3×5+2=17, and for n=3, a_4=4a_3+3=4×17+3=71. Therefore, the correct answer is 71. A value such as 68 can result from using the added term incorrectly in the final step. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

First (a_2=7), then (a_3=15), and finally (a_4=31). In exams, apply the recursive rule step by step.

Under the recursive rule, each new term is the previous term plus twice the term before it. Thus, t_3=5+2(2)=9, t_4=9+2(5)=19, and t_5=19+2(9)=37. Therefore, 37 is correct. A value such as 35 can result from using the wrong earlier terms or missing the coefficient 2. Exam tip: write down t_{n-1} and t_{n-2} separately before substituting their values.

In a recursive rule, use the index of the term currently being found. \(a_2=2(3)+2+1=9\), \(a_3=2(9)+3+1=22\), and \(a_4=2(22)+4+1=49\). Therefore, the correct answer is \(49\). An answer such as \(46\) can result from using an incorrect value of the index in the \(n+1\) part. Exam tip: write the new value of \(n\) at every step before calculating.

By the recursive rule, each new term equals twice the preceding term plus the term before it. Thus, \(c_3=2(3)+1=7\), \(c_4=2(7)+3=17\), and \(c_5=2(17)+7=41\). Therefore, the correct answer is 41. A close distractor such as 39 can result from using an incorrect previous term. Exam tip: write the two preceding terms at every step before calculating the next one.

Work backward using the recursive rule: \(d_4=d_3+8\), \(d_3=d_2+6\), and \(d_2=d_1+4\). Hence, \(d_4=d_1+4+6+8=d_1+18\). Since \(d_4=25\), \(d_1=25-18=7\). If \(d_1=8\), then \(d_4\) would be 26, not 25. Exam tip: when a later term is given in a recursive sequence, apply the rule step by step backward.

Each new term is obtained by doubling the previous term and adding (1). In exams, test the rule on two terms.

The direct answer is Option D: the fifth term. Start with f₁=10. The rule says fₙ=fₙ₋₁−n, so for each new term subtract the new term number from the previous value. Step 1: f₂=f₁−2=10−2=8. Step 2: f₃=f₂−3=8−3=5. Step 3: f₄=f₃−4=5−4=1. Step 4: f₅=f₄−5=1−5=−4. Since −4 is below zero, the fifth term is the first negative term. Option A is wrong because f₂=8 is positive. Option B is wrong because f₃=5 is positive. Option C is wrong because f₄=1 is still positive. Option D is correct because f₅=−4. Notice that the subtracted number is n, not always 1 and not the previous index. A recursive rule must be applied one step at a time, keeping the newly calculated value. Exam cue: calculate in order and stop at the first value less than zero; do not stop at the first small positive value.

The direct answer is A: 26. Use the recursive rule one term at a time, beginning with the given value. We have \\(g_1=2\\). For n=2, add \\(2(2)-1=3\\), so \\(g_2=2+3=5\\). For n=3, add \\(2(3)-1=5\\), so \\(g_3=5+5=10\\). For n=4, add \\(2(4)-1=7\\), so \\(g_4=10+7=17\\). For n=5, add \\(2(5)-1=9\\), so \\(g_5=17+9=26\\). Option A is correct. Option B, 28, is not the result of the stated additions. Option C, 30, is also too large and may come from an arithmetic error. Option D, 32, is not obtained by the recurrence. The important idea is that the rule uses the previous term, not the original first term each time. Check: the added numbers are 3, 5, 7 and 9, and their total is 24; adding 24 to 2 gives 26. Memory cue: calculate every new term from the immediately preceding term.

Under the recursive rule, each term is found by multiplying the previous term by its index. Thus, \(h_2=2\times3=6\), \(h_3=3\times6=18\), and \(h_4=4\times18=72\). Therefore, 72 is correct. An option such as 64 does not follow the required successive multiplications by 2, 3, and 4. Exam tip: write the steps for \(n=2,3,4\) separately to avoid using the wrong index.

In the recursive rule, each term is obtained by adding the square of the current index to the preceding term. Thus, \(j_2=1+2^2=5\), \(j_3=5+3^2=14\), and \(j_4=14+4^2=30\). Therefore, the correct answer is 30. Adding \(1^2\) separately would be incorrect because \(j_1=1\) is already the given initial term. Exam tip: compute recursive sequences step by step from the initial term.