Explicit or general rule

The general term is \(a_n=2^n+2\). Substituting \(n=6\), we get \(a_6=2^6+2=64+2=66\). Although 64 is a close distractor, it is only the value of \(2^6\); the given \(+2\) must also be added. Exam tip: substitute the term number for \(n\), evaluate the power first, and then perform the remaining operations.

Substituting \(n=1,2,3,4\) in \(a_n=3^n+1\) gives \(4,10,28,82\), respectively. Hence, the correct general term is \(a_n=3^n+1\). Although \(a_n=n^3+3\) gives the first term as 4, its second term is 11, not 10. Exam tip: verify a proposed general term using at least the first two or three terms.

Given \(a_n=3^n+1\), substitute \(n=4\): \(a_4=3^4+1=81+1=82\). Option 81 is only the value of \(3^4\); it misses the added 1 in the rule. Exam tip: after substituting the index in a general term, include both the power and the constant term.

(6\cdot2^n) gives (12,24,48,96). In exams check both the base and multiplier in geometric growth.

Given \(a_n=3\cdot2^{n-1}\). Substituting \(n=7\), \(a_7=3\cdot2^{7-1}=3\cdot2^6=3\cdot64=192\). The value 96 can result from incorrectly using \(2^5\). Exam tip: substitute the term number into the exponent \(n-1\) before calculating.

The first term is (3) and each term is multiplied by (2) so (a_n=3\cdot2^{n-1}). In exams check both the first term and ratio.

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.

The governing concept is an explicit rule that generates each term directly from its position n. Test option B at the first four positions. When n = 1, (2n + 3)/5 = 5/5 = 1. When n = 2, it equals 7/5; when n = 3, it equals 9/5; and when n = 4, it equals 11/5. Thus option B reproduces every displayed term exactly. Option A gives 1 for the first term but gives 6/5 for the second, not 7/5. Option C gives 7/3 at n = 1, and option D produces 1, 3, 5, 7. Checking several positions is necessary because matching only the first term cannot establish a valid general rule.

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 direct answer is option B: \(a_n=10n+1\). Start with the meaning of a general term: \(n\) tells the position, so the rule must give 11 when \(n=1\), 21 when \(n=2\), and so on. The sequence increases by 10 each time, so the coefficient of \(n\) is 10. For the first term, \(10(1)=10\), and 1 must be added to obtain 11; hence \(a_n=10n+1\). Checking: at \(n=2\), 21; at \(n=3\), 31. Option A, \(11n\), gives 11 first but 22 second, so its difference is wrong. Option B gives every listed term and is correct. Option C, \(n+10\), gives 11 first but then 12, so it increases by only 1. Option D, \(10n-1\), gives 9 first, not 11. Memory cue: for a linear sequence, use difference times position, then adjust the constant using 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, substitute n=1, 2, 3, and 4: 2¹−1=1, 2²−2=2, 2³−3=5, and 2⁴−4=12. These values reproduce 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 produces square numbers 1, 4, 9, 16. Option D produces 1, 3, 7, 15. Some alternatives agree with the first term, but a valid general rule must generate all the given terms at their correct positions. Only A satisfies that requirement, including the less obvious fourth term.

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 each position n into a candidate and compare the results with the sequence. For option C, n=1 gives 3¹−2(1)=1; n=2 gives 3²−2(2)=9−4=5; n=3 gives 3³−6=27−6=21; and n=4 gives 3⁴−8=81−8=73. Thus aₙ=3ⁿ−2n reproduces every listed term, making option C correct. Option A gives 2 at n=1, so it fails immediately. Option B is linear and gives 4 at n=2. Option D gives 3 at n=1 and 6 at n=2. The term −2n is essential: using only 3ⁿ or using a linear expression cannot account for the stated values.

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.