Recursive rule

The recursive rule forms each new term by adding the previous two terms and 2. Thus, \(a_3=6+3+2=11\), and \(a_4=11+6+2=19\). Therefore, the correct answer is 19. A value such as 17 can result from forgetting to add 2 at one step. Exam tip: calculate \(a_3\) first before finding \(a_4\).

Since 7 is added to each preceding term, the difference between consecutive terms is constantly 7. Hence it is an AP, not a GP. Exam tip: check \(a_{n+1}-a_n\) for a constant difference.

In a recursive rule, the value of n changes at every step. For n=1, \(a_2=2+5(1)-1=6\). For n=2, \(a_3=6+5(2)-1=15\). Then, for n=3, \(a_4=15+5(3)-1=29\). Therefore, the correct answer is 29. A choice such as 27 can result from using an incorrect increment instead of \(14\) in the final step. Exam tip: to find \(a_4\), substitute \(n=1,2,3\) successively.

Apply the recursive rule successively for \(n=1,2,3\): \(a_2=120-(3\times1+2)=115\), \(a_3=115-(3\times2+2)=107\), and \(a_4=107-(3\times3+2)=96\). Hence, the correct answer is \(96\). A value such as \(99\) can result from using an incorrect value of \(n\) in the amount being subtracted. Exam tip: to find \(a_4\) from \(a_1\), apply the recursive rule three times.

Apply the recursive rule successively for n=1, 2, and 3. Thus, \(a_2=2+1^2+1+1=5\), \(a_3=5+2^2+2+1=12\), and \(a_4=12+3^2+3+1=25\). Therefore, the correct answer is 25. A value such as 23 can result from calculating the final increment \(3^2+3+1=13\) incorrectly. Exam tip: To find \(a_4\), apply the rule three times.

Apply the recursive rule first with \(n=1\): \(a_2=a_1+1^2+1+1=2+3=5\). Next, with \(n=2\), \(a_3=a_2+2^2+2+1=5+7=12\). Therefore, the correct answer is 12. The value 14 would result from an incorrect calculation of the second increase. Exam tip: change the value of \(n\) at every recursive step.

The recurrence can be rewritten as \(a_{n+1}=a_n+\frac{a_n}{4}=\frac{5}{4}a_n\). Thus every new term is \(\frac{5}{4}\) times the previous term. Beginning with \(a_1=16\), the second term is \(a_2=16+\frac{16}{4}=16+4=20\). Applying the same rule again gives \(a_3=20+\frac{20}{4}=20+5=25\). Therefore option C is correct.

It is important to use the newly obtained term when calculating the next one. The second step uses 20, not the original 16. This is why the added amount changes from 4 to 5. The sequence begins \(16,20,25\), so the required third term is 25. The rule is a multiplicative growth rule with common factor \(\frac{5}{4}\), although the recursive form is the one asked for.

First simplify the recurrence using like terms. Since aₙ₊₁ = aₙ + 2aₙ, the right side is 3aₙ, so the rule becomes aₙ₊₁ = 3aₙ. This means every term is three times the preceding term. Starting from a₁ = 4, calculate a₂ = 3 × 4 = 12, then a₃ = 3 × 12 = 36, and finally a₄ = 3 × 36 = 108. Therefore option C is correct. The sequence is 4, 12, 36, 108. The other choices do not result from repeated multiplication by 3; they could arise from treating the expression as an addition rule, multiplying by an incorrect factor, or making an error in one of the intermediate terms. Combining like terms is the key step.

Apply the recursive rule step by step: \(a_3=3(2)-2(1)=4\), \(a_4=3(4)-2(2)=8\), and \(a_5=3(8)-2(4)=16\). Therefore, the correct answer is 16. A value such as 15 may result from using an incorrect coefficient or subtraction. Exam tip: in a second-order recurrence, write down both preceding terms before calculating the next term.

(a_3=3\times2-2\times1=4) and (a_4=3\times4-2\times2=8). Exam tip: use both previous terms in the rule.

Using the recursive rule successively for n=1, 2, 3, and 4: a_2=9+2(1)-2=9, a_3=9+2(2)-2=11, a_4=11+2(3)-2=15, and a_5=15+2(4)-2=21. Therefore, the correct answer is 21. A value such as 23 can result from applying the increment 2n-2 incorrectly. Exam tip: while finding each next term, use n equal to the position of the preceding term.

In a recursive rule, substitute a new value of n for each next term. For n=1, a_2=9+2(1)-2=9; for n=2, a_3=9+2(2)-2=11; and for n=3, a_4=11+2(3)-2=15. Therefore, the correct answer is 15. The value 17 would result from incorrectly using an increment of 6 instead of 4 in the last step. Exam tip: To find a_4, apply the rule successively for n=1, 2, and 3.

In option A, each new term is formed from the previous term \(a_{n-1}\), and an initial term \(a_1\) is given, so it is recursive. Rules such as \(a_n=2n+1\) are explicit. Exam tip: look for a previous-term reference.

Using the recursive rule successively, \(a_2=5+\frac{1}{2}\), \(a_3=a_2+\frac{2}{2}\), \(a_4=a_3+\frac{3}{2}\), and \(a_5=a_4+\frac{4}{2}\). Hence, \(a_5=5+\frac{1+2+3+4}{2}=5+5=10\). Therefore, 10 is correct. Choosing 9 would omit the final increase \(\frac{4}{2}\). Exam tip: to find \(a_5\), add the increments for \(n=1\) through \(n=4\).

In the recursive rule, use \(n=1,2,3\) successively to reach the fourth term. Thus, \(a_2=30-\frac12\), \(a_3=30-\frac12-1\), and \(a_4=30-\frac12-1-\frac32=30-3=27\). Therefore, 27 is correct. The value 26 would result from adding the subtractions incorrectly. Exam tip: to find \(a_4\) from \(a_1\), apply the rule three times.

(a_2=1\times2+1=3), (a_3=2\times3+1=7), and (a_4=3\times7+1=22). The correct option should include (22).

In a recursive rule, substitute the appropriate value of n at every step. Thus, a_2=5+2(1)+2=9, a_3=9+2(2)+2=15, a_4=15+2(3)+2=23, and a_5=23+2(4)+2=33. Therefore, 33 is correct. A value such as 31 results from using an incorrect increment in one of the steps. Exam tip: to find a_5, apply the rule successively for n=1 through n=4.

Using n=1 in the recursive rule gives a_2=60-(3×1+1)=56. Then, for n=2, a_3=56-7=49, and for n=3, a_4=49-10=39. Therefore, the correct answer is 39. A value such as 42 can result from calculating the amount to be subtracted incorrectly. Exam tip: to find a_4, apply the rule successively for n=1, 2, and 3.

The governing concept is step-by-step evaluation of an index-dependent recurrence. Begin with a₁ = 3 and substitute n = 1 to obtain a₂ = 2a₁ + 1 + 1 = 2(3) + 2 = 8. Next use n = 2 and the newly calculated term: a₃ = 2a₂ + 2 + 1 = 2(8) + 3 = 19. Finally, use n = 3 to calculate the fourth term: a₄ = 2a₃ + 3 + 1 = 2(19) + 4 = 38 + 4 = 42. Therefore option C is correct. The index changes at every step, so the added quantity is 2, then 3, then 4. Values such as 39, 41, or 45 arise from an arithmetic substitution or indexing error. In particular, the recurrence must be applied three times because the first term is already supplied.

The governing concept is an index-dependent recursive rule. To calculate a_2, use n=1 because the recurrence produces a_{n+1} from a_n: a_2=3a_1−2(1)=3(7)−2=19. To calculate a_3, use n=2 and the newly found value a_2: a_3=3a_2−2(2)=3(19)−4=57−4=53. Therefore option B is correct. The index must change from 1 to 2 in the second application. Reusing n=1, forgetting the factor 3, or subtracting the wrong quantity gives distractor values such as those listed. The recurrence is not a constant-difference rule, so ordinary AP shortcuts cannot replace the stated step-by-step calculation.