Explicit or general rule

This is an arithmetic sequence because each term increases by 3. Checking \(n=1\), \(3(1)+1=4\), and the common difference remains 3. In contrast, \(3n+4\) gives 7 as its first term. Exam tip: verify both the first term and common difference.

The governing concept is selecting an explicit rule by testing the pattern of differences. The first differences are 13−7=6, 23−13=10, and 37−23=14; these increase by 4, so the second difference is constant. A quadratic rule is therefore plausible. Test option A: for n=1, 2(1)^2+5=7; for n=2, 2(2)^2+5=13; for n=3, 2(3)^2+5=23; and for n=4, 2(4)^2+5=37. Thus option A reproduces every given term and is correct. Option B is linear and gives 14 at n=1, option C gives 7 initially but 10 at n=2, and option D gives 7 initially but has constant difference 6, so none matches the sequence.

Given a_n=4n²−n, a_4=4(4²)−4=64−4=60 and a_2=4(2²)−2=16−2=14. Hence, a_4:a_2=60:14=30:7. Option A uses the correct terms but does not simplify the ratio. Exam tip: after finding a ratio, divide both terms by their greatest common factor to write it in simplest form.

Given \(a_n=3n^2+2n-1\). Substituting \(n=2\), \(a_2=3(2)^2+2(2)-1=12+4-1=15\). Similarly, for \(n=3\), \(a_3=3(3)^2+2(3)-1=27+6-1=32\). Hence, option A is correct. Option B has an incorrect value of \(a_2\), while option C has an incorrect value of \(a_3\). Exam tip: substitute the value of \(n\) separately for each term and calculate the square carefully.

The consecutive differences are \(11,17,23\), and their second differences are constant at \(6\). Hence, the sequence has a quadratic general term. Substituting \(n=1,2,3,4\) in \(a_n=3n^2+2n-1\) gives \(4,15,32,55\), respectively. The option \(a_n=11n-7\) matches only the first two terms, not the later terms. Exam tip: verify a proposed general term using at least the first three terms.

The first three terms are (5,14,27) and the sum is (46). In exams write all terms before adding.

Given \(a_n=6n-4\), \(a_{12}=6(12)-4=68\) and \(a_5=6(5)-4=26\). Therefore, \(a_{12}-a_5=68-26=42\). Option 40 may result from an error in finding a term or subtracting the values. Exam tip: substitute the value of \(n\) to find each term separately, then subtract.

Its rule is (a_n=6n-4) and (6n-4=68) gives (n=12). In exams equate the given term to the general term.

Given \(a_n=\frac{2n+3}{5}\). Substituting \(n=6\), \(a_6=\frac{2(6)+3}{5}=\frac{12+3}{5}=\frac{15}{5}=3\). Hence, 3 is the correct answer. \(\frac{13}{5}\) may result from an error in calculating \(2\times6\). Exam tip: substitute the value of \(n\) first, then simplify step by step.

Given \(a_n=10n+1\), \(a_2=10\times2+1=21\) and \(a_6=10\times6+1=61\). Therefore, \(a_2+a_6=21+61=82\). Option 84 would result from calculating one of the terms incorrectly. Exam tip: substitute the value of \(n\) in the general term and find each required term separately.

The first term is (11) and the difference is (10) so (a_n=10n+1). In exams find the constant part from the first term.

Given \(a_n=4n^2+1\), \(a_3=4(3)^2+1=37\) and \(a_4=4(4)^2+1=65\). Therefore, \(a_3+a_4=37+65=102\). An answer such as 100 can result from an error in squaring or addition. Exam tip: Substitute each value of \(n\) separately in the general term before adding.

An explicit rule gives any term directly from its position \(n\). \(a_n=4n-1\) contains only \(n\), whereas the second rule needs the previous term. Exam tip: the presence of \(a_{n-1}\) usually indicates a recursive rule.

The governing idea is to test an explicit formula at the corresponding positive integer positions. For option A, when n=1, 2, 3 and 4, the values are 2^1-1=1, 2^2-2=2, 2^3-3=5, and 2^4-4=12. These are exactly the four displayed terms, so option A is the correct general rule. Option B produces the odd-number pattern 1, 3, 5, 7. Option C gives square numbers 1, 4, 9, 16. Option D gives 1, 3, 7, 15. Although a few formulas may agree with one initial term, the correct rule must reproduce the whole stated sequence; only A does so.

Given \(a_n=3^n-2n\). Substituting \(n=3\), \(a_3=3^3-2(3)=27-6=21\). Therefore, 21 is correct. A result such as 19 would come from an error in evaluating the power or subtraction. Exam tip: substitute the value of \(n\) everywhere before simplifying the expression.

Use the explicit-rule principle: substitute the position n into each candidate and compare the resulting values with the sequence. For option C, n=1 gives 3-2=1; n=2 gives 9-4=5; n=3 gives 27-6=21; and n=4 gives 81-8=73. Thus a_n=3^n-2n reproduces every listed term and is correct. Option A gives 2 at n=1, option B is only linear and gives 4 at n=2, while option D gives 3 at n=1 and 6 at n=2. The subtraction of 2n is essential; using only the power or a simple linear rule does not fit the data.

Given \(a_n=n^3-1\), set \(a_n=124\). Then \(n^3-1=124\), so \(n^3=125=5^3\). Hence \(n=5\), and therefore \(a_5=124\). The close distractor \(a_4\) is incorrect because \(a_4=63\). Exam tip: for expressions of the form \(n^3-1\), add 1 first and identify the perfect cube.

For \(a_n=n^3-1\), we get \(a_1=1^3-1=0\), \(a_2=2^3-1=7\), \(a_3=3^3-1=26\), and \(a_4=4^3-1=63\). Hence, the correct general term is \(a_n=n^3-1\). The nearby distractor \(a_n=n^2-1\) gives \(8\) as its third term, not \(26\). Exam tip: substitute at least the first three values of \(n\) to verify a proposed general 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 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.