nth term

For \(a_4\), substitute \(n=4\) into the given formula: \(a_4=\frac{2(4)^2+1}{4+1}=\frac{2\times16+1}{5}=\frac{33}{5}\). Therefore, \(\frac{33}{5}\) is correct. The option \(7\) is incorrect because \(7=\frac{35}{5}\), whereas the correct numerator is \(2\times16+1=33\). Exam tip: after substituting the term number, evaluate the power first, followed by multiplication and addition.

The terms can be written as \(4-2, 8-2, 16-2, 32-2, 64-2\). The subtracted value is always \(2\), and the nth term of \(4,8,16,32,64\) is \(2^{n+1}\). Hence, the nth term is \(2^{n+1}-2\). Option \(2^{n+2}-6\) gives the first term correctly, but for \(n=2\) it gives \(10\), not \(6\). Exam tip: verify a proposed nth-term formula by substituting \(n=1\) and \(n=2\).

Given \(a_n=n^2+mn+4\), we get \(a_2=2^2+2m+4=8+2m\) and \(a_4=4^2+4m+4=20+4m\). Hence, \(a_2+a_4=28+6m=46\), so \(6m=18\) and \(m=3\). If \(m=4\), the sum would be 52, so it is not correct. Exam tip: substitute the value of \(n\) carefully in every term before combining like terms.

The first differences are \(10,20,30,40\), and their second differences are constantly \(10\). Therefore, the sequence has a quadratic nth term whose \(n^2\) coefficient is \(10/2=5\). Substituting \(n=1,2,3\) in \(a_n=5n^2-5n+3\) gives \(3,13,33\), so it is correct. In option B, the sign of the linear term is incorrect. Exam tip: for a constant second difference, the leading coefficient of a quadratic term is half the second difference.

For \(a_n=5n-3\), \(a_{n+1}-a_n=[5(n+1)-3]-(5n-3)=5\), which is constant for every \(n\). Therefore, it represents an arithmetic progression. The consecutive differences of \(n^2+1\) and \(n(n+1)\) vary, while \(2^n\) is a geometric-type sequence. Exam tip: if \(a_n\) can be written as \(pn+q\), its common difference is \(p\).

Given \(a_n=4n+(-1)^{n+1}\). Substituting \(n=8\), \(a_8=4\times 8+(-1)^{8+1}=32+(-1)^9=32-1=31\). Hence, 31 is correct. The value 33 would result from incorrectly taking \((-1)^9\) as \(+1\); an odd power of \(-1\) is \(-1\). Exam tip: an even power of \(-1\) is \(+1\), while an odd power is \(-1\).

The numerators \(2,5,8,11,\ldots\) form an arithmetic progression whose nth term is \(2+(n-1)\times3=3n-1\). The denominators \(9,16,23,30,\ldots\) also form an arithmetic progression whose nth term is \(9+(n-1)\times7=7n+2\). Hence, the nth term is \(\frac{3n-1}{7n+2}\). Option D has the correct numerator, but for \(n=1\) its denominator is 10 instead of 9. Exam tip: in a sequence of fractions, identify the numerator and denominator patterns separately.

Given \(a_n=2n^3-n\), \(a_4=2(4)^3-4=128-4=124\) and \(a_2=2(2)^3-2=16-2=14\). Therefore, \(a_4-a_2=124-14=110\). A nearby choice such as 108 can result from an error in evaluating the cube or subtracting. Exam tip: find each required term separately before taking their difference.

The first differences are \(13,25,37,49\). Their second differences are all \(12\), so the nth term is quadratic and the coefficient of \(n^2\) is \(12/2=6\). Substituting \(n=1,2,3\) in option B gives \(5,18,43\), respectively; hence \(T_n=6n^2-5n+4\). Although option C gives \(5\) when \(n=1\), it gives \(19\) when \(n=2\), so it is incorrect. Exam tip: when second differences are constant, assume the term has the form \(an^2+bn+c\).

To calculate the difference, replace n by n+1 in the rule. We get \(a_{n+1}=3(n+1)^2-5(n+1)+7\). Expanding gives \(3n^2+6n+3-5n-5+7=3n^2+n+5\). Now subtract \(a_n=3n^2-5n+7\): \(a_{n+1}-a_n=(3n^2+n+5)-(3n^2-5n+7)=6n-2\).

Therefore option C, \(6n-2\), is correct. The square term contributes a changing amount, so the difference is not a constant. Care is needed when subtracting the whole expression: subtracting \(-5n\) contributes \(+5n\), and subtracting 7 contributes \(-7\). Options A and B lose part of this calculation, while option D does not represent the full difference. The supplied answer is correct.

Substitute n=10 into the formula. The first part is \(\frac{10(10+1)}{2}=\frac{10\times11}{2}=55\). This is the sum of the first ten natural numbers, or the tenth triangular number. The formula then adds the fixed constant 3, so \(a_{10}=55+3=58\).

Therefore option A is correct. The constant addition must be included after evaluating the fraction; stopping at 55 would give option B, but that is only the triangular-number part and not the complete term. No further sequence pattern is needed because the explicit formula directly gives the value. The supplied answer and explanation correctly calculate both the variable part and the constant part.

Write the terms as products of consecutive integers: \(12=3\times4\), \(20=4\times5\), \(30=5\times6\), and \(42=6\times7\). Hence, the \(n\)th term is \((n+2)(n+3)\). Expanding gives \(n^2+5n+6\), so option D is correct. Option A gives the first term as 12, but its second term is 19, not 20. Exam tip: For such sequences, factor consecutive terms to identify the pattern in \(n\).

For a term to be positive, \(8n-17>0\). Thus, \(n>\frac{17}{8}\), and the smallest natural number satisfying this is \(3\). Checking confirms that \(a_2=-1\), which is not positive, whereas \(a_3=7\), which is positive. Therefore, the third term is the first positive term. Exam tip: after solving the inequality, choose the smallest natural-number value of \(n\) that satisfies it.

The governing idea is direct substitution into an explicit formula. Replace \(n\) by 7 in both the numerator and denominator: \(a_7=\frac{3(7)-2}{2(7)+1}=\frac{21-2}{14+1}=\frac{19}{15}\). Since 19 and 15 have no common factor, the fraction is already in simplest form. Therefore option C is correct. A common error is to compute the numerator correctly but use an incorrect denominator, producing options A, B, or D. The index must be substituted consistently in every occurrence of \(n\).

The absolute values are 4, 7, 10, 13, 16, which follow the linear pattern \(3n+1\): for \(n=1\), it gives 4, and each next value increases by 3. The signs alternate, beginning with positive. The factor \((-1)^{n+1}\) is positive when \(n=1\), negative when \(n=2\), and continues this alternating pattern. Hence the nth term is \((-1)^{n+1}(3n+1)\), so option A is correct. Option B starts with a negative first term, C omits alternating signs, and D has the wrong magnitude pattern.

Given \(a_n=an^2+bn+1\), putting \(n=1\) gives \(a+b+1=6\), so \(a+b=5\). Putting \(n=2\) gives \(4a+2b+1=17\), so \(2a+b=8\). Solving these equations gives \(a=3\) and \(b=2\). Hence, \(a_5=3(5)^2+2(5)+1=75+10+1=86\). Option 84 would result from omitting the constant term \(+1\). Exam tip: substitute the given term numbers first to form linear equations for the unknown coefficients.

(a_8=180) and (a_3=25), so the difference is (155). In such questions, find both terms separately and subtract.