Explicit or general rule

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 rule followed by simplification of a ratio. For n = 4, a₄ = 6(4²) − 5(4) + 2 = 96 − 20 + 2 = 78. For n = 2, a₂ = 6(2²) − 5(2) + 2 = 24 − 10 + 2 = 16. Hence a₄ : a₂ = 78 : 16. Both terms are divisible by 2, so the simplest form is 39 : 8, making option B correct. Option A is the unsimplified ratio, while C and D result from incorrect substitution or arithmetic.

The governing concept is finding an explicit general term from a sequence. The first differences are 13, 25, and 37; their second differences are both 12, indicating a quadratic rule. Test option A: for n = 1, 6 − 5 + 2 = 3; for n = 2, 24 − 10 + 2 = 16; for n = 3, 54 − 15 + 2 = 41; and for n = 4, 96 − 20 + 2 = 78. It matches every given term, so option A is correct. The linear option C cannot have changing first differences, and B and D fail on one or more 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.

(7n^2+2n-5) gives (4,27,64,115). In exams choose a quadratic rule when second differences are constant.

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.

(4^n+2n-5) gives (1,15,65,259). In exams also check constant subtraction in power-based rules.

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

The governing concept is the explicit rule for an arithmetic sequence. Consecutive terms increase by a constant common difference: 10−6=4, 14−10=4, and 18−14=4. For an arithmetic sequence, the nth term is a_n=a_1+(n−1)d. Substituting a_1=6 and d=4 gives a_n=6+4(n−1)=6+4n−4=4n+2. Therefore, option C is correct. A quick check confirms it: n=1 gives 6, n=2 gives 10, n=3 gives 14, and n=4 gives 18. Option A gives 4 as the first term, option B gives 4 as the first term, and option D has difference 1, so those rules do not fit.

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