Explicit or general rule

For option A, substituting \(n=1,2,3,4\) gives \(6,13,22,33\), respectively. Hence, the correct general term is \(a_n=n^2+4n+1\). The closest distractor, option D, gives the first two terms as \(6,13\), but for \(n=3\) it gives \(20\), not \(22\). Exam tip: test a proposed general term using at least the first three values of \(n\).

Given \(a_n=\frac{n(n+4)}{2}\). For \(a_8\), substitute \(n=8\): \(a_8=\frac{8(8+4)}{2}=\frac{8\times12}{2}=48\). Hence, 48 is the correct option. Values such as 46 or 50 can result from an arithmetic error, but \(8+4=12\). Exam tip: Substitute the required value of \(n\) first, then simplify step by step.

Using (\frac{n(n+4)}{2}) gives the given terms. In exams check fractional terms using small (n) values.

Given \(a_n=2^n+3n-2\). Substituting \(n=4\), we get \(a_4=2^4+3(4)-2=16+12-2=26\). Therefore, option B is correct. Choosing \(28\) usually results from an error in addition or subtraction. Exam tip: evaluate the exponent first, then add or subtract the remaining terms in order.

(2^n+3n-2) gives (3,8,17,36). In exams also check the extra linear part in a power-based rule.

The governing concept is substitution into an explicit sequence rule followed by reduction of a ratio. For n = 4, calculate a₄ = 6(4²) − 5(4) + 2 = 96 − 20 + 2 = 78. For n = 2, calculate a₂ = 6(2²) − 5(2) + 2 = 24 − 10 + 2 = 16. Hence a₄ : a₂ = 78 : 16. The common factor of 78 and 16 is 2, so divide both parts by 2 to obtain 39 : 8. Therefore option B is correct. Option A contains the correct unsimplified values but is not in simplest form. Options C and D result from incorrect substitution or arithmetic and do not represent the given formula.

The governing concept is finding an explicit general term from the observed pattern of a sequence. The first differences are 16−3=13, 41−16=25, and 78−41=37. Their second differences are 25−13=12 and 37−25=12, so a quadratic rule is appropriate. Test option A at each displayed position: when n=1, 6(1)²−5(1)+2=3; when n=2, 24−10+2=16; when n=3, 54−15+2=41; and when n=4, 96−20+2=78. It reproduces every term, so A is correct. Option C is linear and cannot have changing first differences. Options B and D fail when checked against one or more given terms.

Given \(a_n=3\cdot2^n+n\). Substituting \(n=5\), \(a_5=3\cdot2^5+5=3\cdot32+5=96+5=101\). Hence, 101 is the correct option. The value 96 comes from calculating only \(3\cdot2^5\) and incorrectly omitting the final \(+5\). Exam tip: Substitute the term number in every part of the formula before simplifying.

(3\cdot2^n+n) gives (7,14,27,52). In exams check both the power and extra term in rapid growth.

The first six terms are (2,13,24,35,46,57) and the average is (29.5). In exams divide the sum by the number of terms.

The rule is (a_n=11n-9) and (11n-9=134) gives (n=13). In exams equate the given term to the general term.

Given (a_n=n^3+2n^2-n), substitute n=4: (a_4=4^3+2(4^2)-4=64+32-4=92). Therefore, 92 is the correct option. A value such as 94 can result from an error while subtracting the final 4. Exam tip: Substitute the value of n carefully into every term of the rule.

The first differences are \(12,37,89\), and the second differences are \(25,52\). The third difference is constant at \(27\), so an appropriate rule is a cubic polynomial. Substituting \(n=1,2,3,4\) in \(a_n=\frac{9n^3-29n^2+48n-24}{2}\) gives \(2,14,51,140\), respectively. Option B gives \(42\) when \(n=3\), not \(51\). Exam tip: verify a cubic-type rule by substituting at least the first three or four values of \(n\).

(a_2=74) and (a_9=18) so the difference is (56). In exams find both terms carefully in a decreasing formula.

At (n=1) it gives (82) and at (n=2) it gives (74) so (a_n=90-8n). In exams check the first two terms of a decreasing sequence.

For the second term, substitute \(n=2\): \(a_2=7(2)^2+2(2)-5=28+4-5=27\). Similarly, for \(n=3\): \(a_3=7(3)^2+2(3)-5=63+6-5=64\). Hence, option A is correct. The other options result from an error in evaluating the \(n^2\) term or in addition/subtraction. Exam tip: substitute the value of \(n\) separately for each required term and check the arithmetic.

The direct answer is option A, \(a_n=7n^2+2n-5\). Substitute each position carefully. For \(n=1\), \(7(1)^2+2(1)-5=4\). For \(n=2\), \(7(4)+4-5=27\). For \(n=3\), \(7(9)+6-5=64\). For \(n=4\), \(7(16)+8-5=115\). Thus the formula produces every listed term. The first differences are 23, 37, and 51; their differences are 14 and 14, so a quadratic rule is reasonable. Option B, \(4n^2\), gives 4 but then 16 rather than 27. Option C gives 4 at \(n=1\), but 31 at \(n=2\), not 27. Option D gives a linear sequence beginning 4, 27, 50, 73, so it fails after two terms. Memory cue: constant second differences suggest a quadratic rule, but always substitute several values.

Given \(a_n=4^n+2n-5\), substitute \(n=3\): \(a_3=4^3+2(3)-5=64+6-5=65\). Therefore, 65 is correct. A value such as 63 can result from incorrectly evaluating or adding the \(2n\) term. Exam tip: substitute the term number into every part of the rule before simplifying.

The direct answer is option A, \(a_n=4^n+2n-5\). The exponent applies to 4, and then the linear correction \(2n-5\) is added. At \(n=1\), \(4^1+2-5=1\). At \(n=2\), \(4^2+4-5=15\). At \(n=3\), \(4^3+6-5=65\). At \(n=4\), \(4^4+8-5=259\). Hence the formula matches all four terms. Option B gives \(4^1-3=1\), but at \(n=2\) it gives 13, not 15. Option C gives \(2^1+4=6\), so it fails immediately. Option D gives \(1^4=1\), \(2^4=16\), and \(3^4=81\), which do not match. The rapid growth of the terms suggests a power such as \(4^n\), while the small adjustment is checked by substitution. Exam cue: do not ignore the added \(2n-5\) part.

(a_5=109) and (a_1=5) so the sum is (114). In exams recheck the sum before choosing the final option.

The first term is (18) and the difference is (-4), so (a_n=18+(n-1)(-4)=22-4n). Keep the difference negative in a decreasing sequence.

Given \(a_n=4n^2-3n\). To find the fifth term, substitute \(n=5\): \(a_5=4(5)^2-3(5)=4\times25-15=100-15=85\). Hence, the correct answer is 85. The value 100 comes from \(4\times5^2\) alone; the term \(-3n\) must also be subtracted. Exam tip: substitute the term number first, evaluate the power next, and then perform multiplication and subtraction.

Given a_n=6n+1 and a_n=55, set 6n+1=55. Thus, 6n=54 and n=9. Therefore, 55 is the 9th term of the sequence. The 8th term would be 6(8)+1=49, so it is not correct. Exam tip: To find the position of a given term, equate the general term a_n to that value and solve for n.

Direct answer: option D, 6, 12, 20, 30. An explicit rule gives a term directly when we know its position. The first four positions are n = 1, 2, 3, and 4. Substitute each value carefully into aₙ = n² + 3n + 2. For n = 1, a₁ = 1² + 3(1) + 2 = 6. For n = 2, a₂ = 2² + 3(2) + 2 = 4 + 6 + 2 = 12. For n = 3, a₃ = 9 + 9 + 2 = 20. For n = 4, a₄ = 16 + 12 + 2 = 30. Thus D matches every term. A gives 4 as its first value, so it misuses the formula. B gives 5 first and 21 third, so its substitutions are wrong. C has the correct first term but incorrect later calculations. D is the only complete match. Memory cue: for a first-term question, begin with n = 1, not n = 0, unless the question explicitly says otherwise.

The numerator is (n+1) and the denominator is (2n+1), so (a_n=\frac{n+1}{2n+1}). In fractions observe numerator and denominator patterns separately.