Explicit or general rule

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.

Using \(a_n=3^n-1\), we get \(a_1=3-1=2\), \(a_2=9-1=8\), \(a_3=27-1=26\), and \(a_4=81-1=80\). Hence, option A is correct. Although \(a_n=2^n+n\) may look similar, its first term is 3, so it is incorrect. Exam tip: verify a proposed rule with at least the first three terms.

The general term is \(a_n=3^n-1\). Substituting \(n=4\), we get \(a_4=3^4-1=81-1=80\). Option 81 represents only \(3^4\), but the formula also requires subtracting 1. Exam tip: substitute the term number first, then evaluate powers and remaining operations in order.

(5\cdot3^n) gives (15,45,135,405). In exams check both the base and multiplier in geometric growth.

Given \(a_n=4\cdot2^{n-1}\). Substituting \(n=8\), \(a_8=4\cdot2^{8-1}=4\cdot2^7=4\cdot128=512\). Hence, 512 is the correct option. Getting 256 results from an error in the exponent or multiplication. Exam tip: substitute the value of \(n\) into \(n-1\) first, and then simplify.

\(a_n=2n^2-1\) is a quadratic sequence, so its second differences are constant; here they are 4. In \(5n+2\), the first differences are constant instead. Exam tip: look for an \(n^2\) term.

The governing concept is identifying a quadratic explicit rule. The first differences are 16 − 7 = 9, 31 − 16 = 15, and 52 − 31 = 21. Their second differences are 15 − 9 = 6 and 21 − 15 = 6, which suggests a quadratic expression. Test option C: for n = 1, 3(1)² + 4 = 7; for n = 2, 3(2)² + 4 = 16; for n = 3, 3(3)² + 4 = 31; and for n = 4, 3(4)² + 4 = 52. Hence aₙ = 3n² + 4 and option C is correct. Option A is linear and gives 7, 14, 21, ...; option B gives 7, 10, 15, ...; and option D gives 7, 16, 25, ..., so each fails at a later term.

Given a_n=5n^2+n, a_4=5(4)^2+4=80+4=84 and a_2=5(2)^2+2=20+2=22. Hence, a_4:a_2=84:22=42:11. The ratio 11:42 reverses the terms, while 21:11 simplifies only the first term incorrectly. Exam tip: evaluate both terms separately before reducing a ratio by a common factor.

Given \(a_n=2n^2+4n-3\). Substituting \(n=2\), \(a_2=2(2)^2+4(2)-3=8+8-3=13\). Similarly, for \(n=3\), \(a_3=2(3)^2+4(3)-3=18+12-3=27\). Therefore, option A is correct. In option B, the value of \(a_2\) is incorrect. Exam tip: substitute each value of \(n\) separately and calculate the squared term carefully.

The consecutive differences are \(10,14,18\), and their second differences are \(4,4\); therefore, the sequence has a quadratic general term. Substituting \(n=1,2,3,4\) in \(a_n=2n^2+4n-3\) gives \(3,13,27,45\), respectively. \(a_n=10n-7\) matches only the first two terms, not the later terms. Exam tip: verify a proposed general term using at least the first three terms.

Given \(a_n=n^2+4n\), we get \(a_1=1^2+4(1)=5\), \(a_2=2^2+4(2)=12\), and \(a_3=3^2+4(3)=21\). Therefore, the sum of the first three terms is \(5+12+21=38\). Although 39 is a close distractor, the correct addition gives 38. Exam tip: substitute \(n=1,2,3\) first, then add the resulting terms.

\(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant for every term. Hence it is an AP with common difference 5; -2 is only the constant term. Exam tip: in \(pn+q\), \(p\) is the common difference.

Its rule is (a_n=9n-6) and (9n-6=111) gives (n=13). In exams equate the given term to the general term.

Given \(a_n=\frac{3n+2}{4}\), substitute \(n=6\): \(a_6=\frac{3(6)+2}{4}=\frac{18+2}{4}=\frac{20}{4}=5\). The value \(\frac{9}{2}\) may result if the added 2 in the numerator is missed. Exam tip: substitute the given value of \(n\) carefully before simplifying the expression.