Explicit or general rule

Substituting \(n=1,2,3,4\) in \(a_n=2^n+n\) gives \(3,6,11,20\), respectively. Hence, the correct general term is \(a_n=2^n+n\). Although \(a_n=n^2+2\) matches the first three terms, it gives \(18\), not \(20\), when \(n=4\). Exam tip: verify a proposed rule using at least four given terms.

Given \(a_n=3^n-n\). Substituting \(n=3\), we get \(a_3=3^3-3=27-3=24\). Hence, the correct answer is 24. The option 27 is only the value of \(3^3\); subtracting 3 is also required. Exam tip: substitute the value of \(n\) carefully in every part of the general term before calculating.

Evaluate each rule using the term number n, beginning with n = 1. For option D, a₁ = 3¹ − 1 = 2, a₂ = 3² − 2 = 7, a₃ = 3³ − 3 = 24, and a₄ = 3⁴ − 4 = 77. It therefore matches every listed term, so option D is correct. The sequence grows mainly by powers of 3, with the index subtracted. Option A gives 4, 10, 28, 82; option B gives 3, 6, 11, 20; and option C gives 2, 5, 8, 11. Although option A may look similar because it uses a power of 3, its added constant does not fit the second term. Substitution at several indices is the safest verification method for an explicit rule.

Putting (n=5) in (2n^2+2n=60) gives (60). In exams, you can also check quickly by substituting options.

The first differences are 8, 12, and 16, so the sequence is not arithmetic. The second differences are 4 and 4, which suggests a quadratic rule. Substitute n = 1, 2, 3, and 4 into option B: a₁ = 2(1)² + 2(1) = 4; a₂ = 2(2)² + 2(2) = 8 + 4 = 12; a₃ = 18 + 6 = 24; and a₄ = 32 + 8 = 40. Thus option B reproduces the complete given sequence. Option A gives 4, 8, 12, 16 and is linear. Option C gives 4, 10, 18, 28, while option D gives 2, 8, 18, 32. Both fail after the first term, so B is the unique correct general term.

For \(a_n=3n+2\), increasing \(n\) by 1 increases every term by 3, so the common difference is constant and it is an AP. In \(n^2+2\), the differences change. Exam tip: check the difference between consecutive terms.

Answer: B, aₙ = n² + 3n + 1. An explicit formula gives a term directly from its position n. Test option B: for n = 1, 1 + 3 + 1 = 5; for n = 2, 4 + 6 + 1 = 11; for n = 3, 9 + 9 + 1 = 19; and for n = 4, 16 + 12 + 1 = 29. Thus it matches every displayed term. Option A gives 5, 8, 13, 20, so it fails from the second term. Option C gives 5, 11, 21, 35, so it fails at the third term. Option D gives 5, 14, 29, 50, so it also fails at the second term. The successive differences of the sequence are 6, 8, and 10, showing a quadratic pattern with constant second difference 2. Memory cue: substitute n = 1, 2, 3 into a proposed explicit rule before accepting it.

The governing concept is using an explicit sequence rule to identify the position of a specified term. Set the general term equal to the required value: 6n−4=62. Add 4 to both sides to obtain 6n=66, and divide by 6 to obtain n=11. Hence 62 is the 11th term, so option C is correct. Verification is immediate: a_11=6(11)−4=66−4=62. The other choices would produce a_9=50, a_10=56, and a_12=68, respectively, so they cannot equal 62. The important method is to solve for the index n rather than substitute the given value into the wrong part of the formula.

The governing concept is the general term of an arithmetic sequence. The consecutive differences are 13−7=6, 19−13=6, and 25−19=6, so the common difference is 6. For an arithmetic sequence, a_n=a_1+(n−1)d. Substituting a_1=7 and d=6 gives a_n=7+6(n−1)=7+6n−6=6n+1. Therefore option A is correct. Checking n=1 gives 7, n=2 gives 13, and n=4 gives 25, confirming the rule. Option C gives 5 for the first term, option B has an incorrect coefficient and fails at n=2, and option D increases by only 1 rather than by 6.

Direct answer: aₙ = 4n + 2, so option C is correct. Subtract consecutive terms: 10−6 = 4, 14−10 = 4, and 18−14 = 4. The constant difference shows that this is an arithmetic sequence with first term 6 and common difference 4. Apply the formula aₙ = a₁ + (n−1)d: aₙ = 6 + 4(n−1) = 6 + 4n − 4 = 4n + 2. Checking gives 6 when n = 1, 10 when n = 2, 14 when n = 3, and 18 when n = 4. Option A gives 4 as the first term, so it is too small by 2. Option B also gives 4 when n = 1 and therefore fails immediately. Option C gives both the correct first term and the correct increase of 4. Option D begins at 6 but increases by 1, producing 7 next instead of 10. Common confusion: the coefficient of n gives the common difference, while the constant term adjusts the first term.

For \(a_n=5n-2\), \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant, so it is an arithmetic progression. For \(n^2+1\), the differences change. Exam tip: check consecutive-term differences.

Direct answer: aₙ = 7n + 2, so option D is correct. First calculate the consecutive differences: 16−9 = 7, 23−16 = 7, and 30−23 = 7. Thus the sequence is arithmetic with first term a₁ = 9 and common difference d = 7. The standard formula is aₙ = a₁ + (n−1)d. Substitution gives aₙ = 9 + 7(n−1) = 9 + 7n − 7 = 7n + 2. Check: n = 1 gives 9, n = 2 gives 16, n = 3 gives 23, and n = 4 gives 30. Option A gives 9 at n = 1 but its difference is 9, so it would continue 18, 27, 36. Option B gives 5 at n = 1. Option C gives the correct first term 9 but has difference 2, not 7. Option D satisfies both required features, so it is the only correct rule. Memory cue: in an expression pn+q for an arithmetic sequence, p is the common difference; then use n = 1 to determine q.

Given \(a_n=60-5n\), substitute \(n=9\) to find the ninth term: \(a_9=60-5(9)=60-45=15\). Therefore, 15 is correct. The value 20 could result from incorrectly calculating \(5\times9\) as 40. Exam tip: substitute the term number for \(n\) carefully in a general-term formula.

At (n=1) it gives (47) and at (n=2) it gives (43) so (a_n=51-4n). In exams match the first term in a decreasing sequence.

The governing concept is using an explicit rule to identify the position of a specified term. We are given a_n=5n+4 and want a_n=64, so set the formula equal to 64: 5n+4=64. Subtracting 4 from both sides gives 5n=60, and dividing by 5 gives n=12. Therefore, option A is correct: the twelfth term is 64. Substitution confirms it because a_12=5(12)+4=60+4=64. The other choices fail the formula: n=11 gives 59, n=13 gives 69, and n=10 gives 54. The essential method is to equate the explicit expression with the required value and solve for the positive integer term number.

Its rule is (a_n=3n-1) and (3n-1=38) gives (n=13). In exams form the general term first to find the term number.

The direct answer is option B, \(a_n=(n+2)^2\). Write the terms as squares: \(9=3^2\), \(16=4^2\), \(25=5^2\), and \(36=6^2\). The square roots are 3, 4, 5, 6, which equal \(n+2\) when \(n=1,2,3,4\). Therefore the general term is \((n+2)^2\). Checking gives \(3^2=9\), \(4^2=16\), \(5^2=25\), and \(6^2=36\). Option A, \(n^2\), gives 1, 4, 9, 16, so it starts three square numbers too early. Option C gives 3, 6, 11, 18, so adding 2 is not the same as shifting the square. Option D gives 3, 12, 27, 48, which is much larger and does not match. Important distinction: \((n+2)^2\) means square after shifting, not \(n^2+2\).

Given \(a_n=(n+2)^2\). Substituting \(n=5\), we get \(a_5=(5+2)^2=7^2=49\). Therefore, 49 is correct. The value 36 would result from using 6 before squaring, which is not the correct substitution here. Exam tip: substitute the value of \(n\) first, simplify the bracket, and then square it.

Given \(a_n=n^2+2n\), substitute \(n=6\): \(a_6=6^2+2(6)=36+12=48\). Therefore, the correct answer is \(48\). A value such as \(46\) may result from not evaluating or adding the squared term and the \(2n\) term correctly. Exam tip: substitute the term number for every occurrence of \(n\) in the expression.

Its rule is (a_n=n^2) and (n^2=81) gives (n=9). In exams identify the term number of square numbers.

Given \(a_n=\frac{n(n+1)}{2}+2\), substitute \(n=6\): \(a_6=\frac{6(6+1)}{2}+2=\frac{42}{2}+2=21+2=23\). Hence, the correct answer is 23. The value 21 is only \(\frac{6\times7}{2}\); the given \(+2\) must still be added. Exam tip: when evaluating a general term, substitute the value of \(n\) carefully and perform the operations in order.

(\frac{n(n+1)}{2}+2) gives (3,5,8,12). In exams think of triangular numbers when differences are (2,3,4).

The first differences are \(5,7,9\), increasing by 2 each time, so the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=n^2+n-2\) gives \(0,5,12,21\), respectively. Option A gives \(3\) as the second term, so it does not fit. Exam tip: verify a proposed general term by checking at least the first four values of \(n\).

Set \(n^2+n-2=70\). This gives \(n^2+n-72=0\), or \((n+9)(n-8)=0\). The roots are \(n=8\) and \(n=-9\), but a term number must be positive, so \(n=8\) is correct. For example, \(n=7\) gives 54, not 70. Exam tip: reject a negative root when it cannot represent a term number.

The correct rule is \(a_n=2^n+2\). Substituting \(n=1,2,3,4\) gives \(4,6,10,18\), respectively. Although \(a_n=2n+2\) gives the first two terms as 4 and 6, its third term is 8, not 10. Exam tip: verify an explicit rule using at least the first three terms.