Explicit or general rule

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

The correct rule is \(a_n=2n^3+n\). Checking it: for \(n=1\), \(a_1=2(1)^3+1=3\); for \(n=2\), \(a_2=2(2)^3+2=18\); and for \(n=3\), \(a_3=2(3)^3+3=57\). Thus, it matches the given sequence. In contrast, \(3n^2\) gives 12 as the second term, not 18. Exam tip: substitute at least the first two or three values of \(n\) to verify a proposed general term.

(a_4=56) and (a_7=38) so the sum is (94). In exams find both terms carefully in a decreasing formula.

At (n=1) it gives (74) and at (n=2) it gives (68) so (a_n=80-6n). In exams check the first two terms of a decreasing sequence.

To find \(a_4\), substitute \(n=4\) in the general-term formula: \(a_4=\frac{4(2\times4+1)}{3}=\frac{4\times9}{3}=12\). Therefore, the correct answer is 12. Values such as 10 or 11 may result from evaluating \(2n+1\) incorrectly. Exam tip: substitute the required term number carefully before simplifying.

To verify an explicit rule, substitute consecutive values of n and simplify each expression carefully. For option A, n=1 gives 1(2+1)/3=1; n=2 gives 2(4+1)/3=10/3; n=3 gives 3(6+1)/3=7; and n=4 gives 4(8+1)/3=12. This agrees with every displayed term, so option A is correct. Option B generates 1, 3, 6, 10, the triangular-number sequence. Option C generates 1, 3, 5, 7, while option D generates 2, 7, 15, 26. The fractional second term is a useful discriminator: a rule that ignores the denominator or changes the growth pattern cannot be accepted, even if its first term happens to be 1.

Given \(a_n=4^n-n\). Substituting \(n=3\), we get \(a_3=4^3-3=64-3=61\). Therefore, 61 is the correct option. The value 60 would result from subtracting 4 from 64, but the formula requires subtracting \(n=3\). Exam tip: substitute the term number first, then evaluate the exponent carefully.

The correct rule is \(a_n=2n^2+3n\). Substituting \(n=1,2,3,4\) gives \(5,14,27,44\), respectively. Option A gives the first term as 5, but its second term is \(12\), so it does not fit the sequence. Exam tip: Always test a proposed general term for at least the first three terms.

The governing concept is an explicit or general rule: a formula must produce the term directly from its position n. Rewrite the sequence as 2³, 3³, 4³, 5³, ... . The first displayed term corresponds to n = 1, so its cube base is n + 1; therefore the rule is aₙ = (n + 1)³. Checking confirms it: for n = 1, a₁ = 2³ = 8; for n = 2, a₂ = 3³ = 27; for n = 3, a₃ = 4³ = 64; and for n = 4, a₄ = 5³ = 125. Thus option B is correct. Option A starts with 1³, option C gives 2 at n = 1, and option D starts with 3³, so those alternatives do not match the sequence.

The governing concept is substitution into an explicit sequence rule, followed by subtraction of two terms. For the eighth term, substitute n = 8: a₈ = 9(8) − 4 = 72 − 4 = 68. For the third term, substitute n = 3: a₃ = 9(3) − 4 = 27 − 4 = 23. Therefore a₈ − a₃ = 68 − 23 = 45, so option B is correct. A useful shortcut is that the constant term cancels: (9 × 8 − 4) − (9 × 3 − 4) = 9(8 − 3) = 45. Option A could result from an arithmetic error, while options C and D do not follow from the required evaluations. Both terms must be calculated using their correct indices before finding the difference.

The governing concept is finding an explicit rule for an arithmetic sequence. Consecutive terms increase by a constant difference: 13 − 8 = 5, 18 − 13 = 5, and 23 − 18 = 5. An arithmetic sequence has aₙ = a₁ + (n − 1)d, so here aₙ = 8 + (n − 1)5 = 8 + 5n − 5 = 5n + 3. Thus option B is correct. Verification is important: at n = 1, the formula gives 5(1) + 3 = 8; at n = 2 it gives 13; at n = 3 it gives 18; and at n = 4 it gives 23. Option A gives 8, 16, 24, ... and has the wrong difference. Option C begins with 2, and option D has difference 3, so neither describes the given sequence.

\(a_n=5n-2\) has the form \(pn+q\), so \(a_{n+1}-a_n=5\) is constant. For \(n^2+1\), successive differences change. Exam tip: check whether the common difference is constant.

The direct answer is option A: \(a_n=9n+5\). Compare consecutive terms: \(23-14=9\), \(32-23=9\), and \(41-32=9\). Thus this is an arithmetic sequence with common difference 9. A linear rule has the form \(a_n=9n+c\). Using the first term, \(9(1)+c=14\), so \(c=5\); therefore \(a_n=9n+5\). Checking gives 14, 23, 32, and 41 in order. Option A is correct. Option B, \(14n\), gives 14 first but 28 second, so it does not preserve the difference. Option C, \(9n-5\), gives 4 first, not 14. Option D, \(5n+9\), gives the first term 14 but has difference 5, so its second term is 19. Exam cue: identify the common difference first, then use the first term to find the constant.

At (n=1) it gives (70) and at (n=2) it gives (64) so (a_n=76-6n). In exams match the first term in a decreasing sequence.

The governing concept is finding the position of a term from an explicit formula. Set the given rule equal to the target value: 4n + 9 = 73. Subtract 9 from both sides to obtain 4n = 64, then divide by 4 to get n = 16. Therefore 73 is the sixteenth term, and option C is correct. Direct checking gives a₁₆ = 4(16) + 9 = 64 + 9 = 73. The other options produce different values: a₁₄ = 65, a₁₅ = 69, and a₁₇ = 77. A common error is to subtract 9 correctly but divide incorrectly, or to confuse the value of a term with its position. Here n is the position, while 73 is the term value.

Its rule is (a_n=7n-2) and (7n-2=61) gives (n=9). In exams form the general term first to find the term number.

The direct answer is option B, \(a_n=(n+3)^2\). The terms are consecutive squares: \(16=4^2\), \(25=5^2\), \(36=6^2\), and \(49=7^2\). For positions \(n=1,2,3,4\), the square roots are respectively \(n+3\), so the rule is \((n+3)^2\). Substitution confirms it: \((1+3)^2=16\), \((2+3)^2=25\), \((3+3)^2=36\), and \((4+3)^2=49\). Option A, \(n^2\), gives 1, 4, 9, 16, so it begins too early. Option C, \(n^2+3\), gives 4, 7, 12, 19; adding 3 after squaring does not create the required shifted squares. Option D, \(4n^2\), gives 4, 16, 36, 64, so it has a different pattern. The key is to find the square root first, then express that root using \(n\).

The general term is \(a_n=(n+3)^2\). Substituting \(n=5\), we get \(a_5=(5+3)^2=8^2=64\). Hence, 64 is the correct answer. The value 81 would result only if the quantity inside the bracket were 9. Exam tip: evaluate the expression inside brackets before squaring it.

For \(a_n=n^2+3n+2\), substituting \(n=1,2,3,4\) gives 6, 12, 20, and 30 respectively. Hence, option A is correct. Option B gives 12 as its second term, but its third term is 18, not 20. Exam tip: verify a proposed general term by checking at least the first three terms.

In option A, \(a_{n+1}-a_n=[3(n+1)-1]-(3n-1)=3\), which is constant, so it is an arithmetic progression. For \(n^2+1\), the differences change. Exam tip: check consecutive-term differences.

Adding (3) to triangular numbers gives (4,6,9,13). In exams think of triangular numbers when differences are (2,3,4).

Substitute \(n=7\): \(a_7=\frac{7(7+1)}{2}+3=\frac{7\times8}{2}+3=28+3=31\). Hence, 31 is correct. The close distractor 28 is only the value of \(\frac{7\times8}{2}\); the given \(+3\) must still be added. Exam tip: Substitute the required index for \(n\) and simplify step by step.

In option A, substituting \(n=1,2,3,4\) gives \(1,6,13,22\), respectively. Therefore, the general term is \(a_n=n^2+2n-2\). Option C matches only the first two terms; at \(n=3\), it gives 11 instead of 13. Exam tip: verify a proposed general term using at least three or four terms.

Putting (n=9) gives (97), while (n=8) gives (78), so none of the options is correct. In exams check options by direct substitution.

Given \(a_n=2^n+3\), substitute \(n=6\): \(a_6=2^6+3=64+3=67\). Option 65 could result from an addition error; the constant 3 must be added to 64. Exam tip: to find a term from an explicit rule, substitute the term number first and evaluate the exponent carefully.