Explicit or general rule

The first differences are \(7,11,15\), and the second difference is constant at \(4\), so the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=2n^2+n\) gives \(3,10,21,36\), respectively. Option A gives the first term as \(3\), but its second term is \(8\), not \(10\). Exam tip: verify a proposed general term using at least the first three terms.

In option A, the first differences are \(3,5,7,9\), so the second difference is constantly \(2\). A constant second difference indicates a quadratic rule. Option B has constant first differences and is linear. Exam tip: make a difference table to classify a sequence.

The correct rule is \(a_n=3n^2+2n\). Checking it gives \(a_1=3+2=5\), \(a_2=12+4=16\), and \(a_3=27+6=33\). The linear rule \(11n-6\) may match the first two terms, but for \(n=3\) it gives 27, not 33. In an exam, test an explicit rule with at least three terms.

(a_5=98) and (a_2=11), so the difference is (87). In exams, find both terms correctly before subtracting.

The first differences are \(7,13,19\), and the second differences are constant: \(6,6\). Hence, the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=3n^2-2n+1\) gives \(2,9,22,41\), respectively. Option C matches the first two terms but gives \(18\), not \(22\), when \(n=3\). Exam tip: constant second differences usually indicate a rule of the form \(an^2+bn+c\).

(a_7=\frac{7\times10}{2}+2=37). In exams, simplify the fraction first and then add the constant.

The terms \(4,9,16,25\) are respectively \(2^2,3^2,4^2,5^2\). When \(n=1\), the number being squared is \(2=n+1\); hence the general term is \(a_n=(n+1)^2\). Although \(a_n=n^2+3\) gives 4 for the first term, it gives 7, not 9, when \(n=2\). Exam tip: verify a general term by substituting \(n=1\) and \(n=2\).

Given \(a_n=(2n-1)^2\). Substituting \(n=5\), we get \(a_5=(2\times5-1)^2=(10-1)^2=9^2=81\). Hence, 81 is the correct option. The value 64 would require the expression inside the square to be 8, which it is not. Exam tip: substitute the value of \(n\) first, and then evaluate the exponent.

The direct answer is option A: \(a_n=4n^2-n+1\). First differences are 11, 19, and 27; the second differences are \(8\) and \(8\). Constant second differences indicate a quadratic rule, so a squared term should be tested. Substitute each position into option A: for \(n=1\), \(4-1+1=4\); for \(n=2\), \(16-2+1=15\); for \(n=3\), \(36-3+1=34\); for \(n=4\), \(64-4+1=61\). Thus A exactly fits. Option B, \(3n^2+1\), gives 4 first but 13 second. Option C, \(n^2+10n-7\), gives 4 first but 17 second. Option D, \(11n-7\), is linear and gives 4 first but 15 second, so it cannot match the changing differences. Memory cue: constant second difference usually signals a quadratic sequence.

\(a_{n+1}-a_n=[12-4(n+1)]-(12-4n)=-4\), which is constant for every \(n\). Hence it is an arithmetic progression with common difference \(-4\), not \(4\). Exam tip: check the difference of consecutive terms.

At (n=1) it gives (120), and at (n=2) it gives (112), so (a_n=128-8n). In exams, keep the decreasing difference negative.

The governing concept is using an explicit term rule in reverse to find the position of a specified value. Set the general expression equal to 146: 9n + 2 = 146. Subtract 2 from both sides to obtain 9n = 144, and divide by 9 to get n = 16. Hence the value 146 is the 16th term, so option C is correct. Verification gives a₁₆ = 9(16) + 2 = 144 + 2 = 146. Option B gives 137, while option D gives 155; option A gives 128. The essential step is solving the equation for the term number rather than substituting a guessed position into the sequence.

Its rule is (a_n=9n-3), and (9n-3=150) gives (n=17), so the correct term is the (17)th term.

Take \(n=1,2,3,4\). The values of \(n^2\) are \(1,4,9,16\); adding 1 gives \(2,5,10,17\). Hence, \(a_n=n^2+1\). The close distractor \(a_n=n^2+n\) gives 6 when \(n=2\), not 5. Exam tip: verify a proposed general term using at least the first three terms.

Given \(a_n=n^2+1\), set \(a_n=101\). Then \(n^2+1=101\), so \(n^2=100\) and \(n=10\). Hence, 101 is the 10th term of the sequence. Although the equation also gives \(n=-10\), a term number must be a positive integer, so it is rejected. Exam tip: move the constant term first, then take the square root of the resulting perfect square.

Given \(a_n=2^n+3n\), substitute \(n=5\): \(a_5=2^5+3(5)=32+15=47\). Option 43 may result from adding only 3 instead of evaluating \(3n\), which is incorrect. Exam tip: after substituting the value of \(n\), evaluate powers and multiplication before addition.

An explicit rule should produce every listed term when the position n is replaced by 1, 2, 3, and so on. The first term is obtained with n=1, the second with n=2, and the fourth with n=4. Testing several terms is important because a rule may match one value by coincidence. Here option A combines an exponential part \\(2^n\\) with a linear part \\(3n\\).

For option A, at n=1 we get \\(2^1+3(1)=5\\); at n=2, \\(2^2+3(2)=10\\); at n=3, \\(2^3+3(3)=17\\); and at n=4, \\(2^4+3(4)=28\\). These are exactly the given terms, so option A is correct. The other rules fail at one or more early positions and therefore cannot be the stated explicit rule.

Given \(a_n=3^n-2n+1\), substitute \(n=4\): \(a_4=3^4-2(4)+1=81-8+1=74\). Therefore, 74 is the correct answer. The value 72 would result from omitting the final \(+1\), so it is incorrect. Exam tip: After substituting the term number, evaluate the exponent and multiplication in brackets carefully.

Substituting \(n=1,2,3,4\) into \(a_n=3^n-2n+1\) gives \(2,6,22,74\), respectively. Hence, it is the general term. The close distractor \(2^n+n\) gives 3 as its first term, not 2. Exam tip: verify a proposed general term by checking it for at least the first three values of \(n\).

Substituting \(n=6\), \(a_6=5\cdot2^{6-1}-3=5\cdot2^5-3=5\cdot32-3=157\). Hence, 157 is correct. The close distractor 153 can result from mishandling the final \(-3\) or making an arithmetic error. Exam tip: substitute the term number into the exponent \(n-1\) first, then simplify step by step.

The direct answer is option A: \(a_n=5\cdot2^{n-1}-3\). To test a general term, substitute the term number \(n=1,2,3,4\) and compare the results with 2, 7, 17, 37. For option A, when \(n=1\), \(5\cdot2^0-3=5-3=2\). When \(n=2\), \(5\cdot2^1-3=10-3=7\). When \(n=3\), \(5\cdot2^2-3=20-3=17\). When \(n=4\), \(5\cdot2^3-3=40-3=37\). Thus it matches every displayed term. Option B, \(2^n+1\), gives 3 for the first term, not 2, and then gives 5, 9, 17, so it does not match the sequence. Option C, \(5n-3\), gives 2, 7, 12, 17; it is an arithmetic rule and misses the fourth term. Option D, \(3\cdot2^n-4\), gives 2, 8, 20, 44, so only the first term happens to match. The sequence is not arithmetic because its differences are 5, 10, and 20; these differences double. The formula in A reflects doubling followed by subtracting 3. The safest exam method is to check at least the first four terms, especially when exponential and linear options are mixed.

Using a_n=3n^2+n-2: a_1=3(1)^2+1-2=2, a_2=3(2)^2+2-2=12, and a_3=3(3)^2+3-2=28. Therefore, the sum of the first three terms is 2+12+28=42. Option 40 is incorrect because the third term is 28, not 26. Exam tip: substitute each value of n separately before adding the terms.

For \(a_n=7-4n\), \(a_{n+1}-a_n=[7-4(n+1)]-(7-4n)=-4\), which is constant and non-zero for every \(n\). Hence it is an AP. In option B, the difference changes with \(n\). Exam tip: an AP has an nth-term rule linear in \(n\).

Substituting \(n=1,2,3,4\) into \(a_n=2n^3-n\) gives \(1,14,51,124\), respectively. Hence, option B is correct. The close distractor \(n^3+1\) gives 9 when \(n=2\), not the required second term 14. Exam tip: test an explicit rule using at least the first three values of \(n\).

For \(a_n=n^3+n^2\), substituting \(n=1,2,3,4\) gives \(2,12,36,80\), respectively. Hence, this is the general term of the sequence. The close distractor \(4n^2-2n\) gives 2 and 12 initially, but for \(n=3\) it gives 30, not 36. Exam tip: verify a proposed general term using at least the first three values of \(n\).