Explicit or general rule

(a_4=56) and (a_7=38) so the sum is (94). In exams find both terms carefully in a decreasing formula.

At (n=1) it gives (74) and at (n=2) it gives (68) so (a_n=80-6n). In exams check the first two terms of a decreasing sequence.

To find \(a_4\), substitute \(n=4\) in the general-term formula: \(a_4=\frac{4(2\times4+1)}{3}=\frac{4\times9}{3}=12\). Therefore, the correct answer is 12. Values such as 10 or 11 may result from evaluating \(2n+1\) incorrectly. Exam tip: substitute the required term number carefully before simplifying.

To verify an explicit rule, substitute consecutive values of n and simplify carefully. For option A, n=1 gives 1(3)/3=1; n=2 gives 2(5)/3=10/3; n=3 gives 3(7)/3=7; and n=4 gives 4(9)/3=12. This matches every displayed term, so A is correct. Option B gives 1, 3, 6, 10, the triangular-number sequence. Option C gives 1, 3, 5, 7, and option D gives 2, 7, 15, 26. Matching fractional values is important here because a rule that resembles the pattern but ignores the denominator is not valid.

Given \(a_n=4^n-n\). Substituting \(n=3\), we get \(a_3=4^3-3=64-3=61\). Therefore, 61 is the correct option. The value 60 would result from subtracting 4 from 64, but the formula requires subtracting \(n=3\). Exam tip: substitute the term number first, then evaluate the exponent carefully.

The correct rule is \(a_n=2n^2+3n\). Substituting \(n=1,2,3,4\) gives \(5,14,27,44\), respectively. Option A gives the first term as 5, but its second term is \(12\), so it does not fit the sequence. Exam tip: Always test a proposed general term for at least the first three terms.

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.

The governing concept is evaluating an explicit sequence rule at n=1, 2, 3 and 4. Substitute each position into a_n=n^2+3n+2: a_1=1^2+3(1)+2=6, a_2=2^2+3(2)+2=12, a_3=3^2+3(3)+2=20, and a_4=4^2+3(4)+2=30. Therefore the first four terms are 6, 12, 20 and 30, so option D is correct. Option A appears to omit part of the expression, option B does not result from the stated quadratic rule, and option C gives incorrect substitutions. The indexing starts at 1, not 0, because the first term is conventionally a_1.

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.

The rule is a_n=n(n+2). Substituting n=1 gives 1(1+2)=3; n=2 gives 2(2+2)=8; n=3 gives 3(3+2)=15; and n=4 gives 4(4+2)=24. Therefore, the first four terms are (3, 8, 15, 24), so option C is correct. Option A begins with 2, but the correct first term for n=1 must be 3. Exam tip: For a general-term question, substitute n=1, 2, 3, and so on in order.

For the fifth term, substitute \(n=5\): \(a_5=5\cdot2^{5-1}=5\cdot2^4=5\cdot16=80\). Therefore, 80 is correct. The value 160 would result from incorrectly using \(2^5\) instead of \(2^{n-1}\). Exam tip: Read the exponent carefully before substituting the term number.

This is (3^2,4^2,5^2,6^2,\ldots), so (a_n=(n+2)^2). In square sequences relate the base number to (n).

The sign alternates and the magnitude is (n), so (a_n=(-1)^{n+1}n). In alternating signs choose the power of ((-1)^n) carefully.

Substitute n=4 into both rules. \(a_4=2(4)+1=9\) and \(b_4=4^2-1=15\). Therefore, \(a_4+b_4=9+15=24\). A result such as 26 comes from evaluating one of the terms incorrectly. Exam tip: find each sequence term separately before adding them.

Its rule is (a_n=n^2+3n+1), so (a_8=64+24+1=89). First identify the general rule and then find the term.

Taking the term number as \(n=1,2,3,\ldots\), the rule \(a_n=n^2+2n-1\) gives \(a_1=2\), \(a_2=7\), \(a_3=14\), and \(a_4=23\). Hence, option A is correct. The closest distractor, \(n^2+n\), gives the second term as \(6\), not \(7\). Exam tip: when first differences increase regularly, such as \(5,7,9\), test a quadratic rule.

The general term is (a_n=4n+4), and (4n+4=50) gives (n=\frac{23}{2}). If the position is not a natural number, the given number is not a term.

These are squares of odd numbers (1^2,3^2,5^2,7^2,\ldots), so (a_n=(2n-1)^2). In squares observe the base pattern.