nth term

Look at the terms by separating a power of 2 from the remaining constant: 5 is \(2^2+1\), 9 is \(2^3+1\), 17 is \(2^4+1\), 33 is \(2^5+1\), and 65 is \(2^6+1\). The exponent starts at 2 when the position is 1, so the exponent in the nth term is \(n+1\). Hence the general term is \(a_n=2^{n+1}+1\).

Checking the formula confirms it: for n=1 it gives \(2^2+1=5\), and for n=5 it gives \(2^6+1=65\). Therefore option A is correct. Option B has the right type of expression but gives 5 for the first term, while option D starts with a different power and gives an incorrect value. Recognizing the doubling pattern and the fixed added 1 is the key step.

(a_9=108) and (a_4=28), so the difference is (88), not (80). Open brackets carefully in product rules.

The sequence is formed by squaring the positive counting numbers: the first term is 1 squared, the second is 2 squared, the third is 3 squared, and so on. Thus its general term is \\(a_n=n^2\\). To find the position of 196, solve \\(n^2=196\\). Since \\(14^2=196\\) and the term number is positive, \\(n=14\\). Therefore, 196 is the 14th term, so option A is correct. The other listed positions would produce different squares.

The pattern can also be checked directly: the terms are 1, 4, 9, 16, 25, ..., which are the squares of 1, 2, 3, 4, 5, ... . Continuing this rule, the number obtained from the 14th position is \\(14^2=196\\). It is not necessary to count every term individually. Taking the positive square root gives the term number because sequence positions are positive integers. The supplied answer A and its reasoning are accurate.

Given \(a_n=12-5n\), \(a_3=12-5(3)=-3\) and \(a_8=12-5(8)=-28\). Therefore, \(a_3+a_8=-3+(-28)=-31\). The value \(-29\) can result from an error while substituting the term number or handling the negative sign. Exam tip: find each required term separately before adding them.

Given \(a_n=\frac{n(n+1)}{2}\). Substituting \(n=12\), \(a_{12}=\frac{12(12+1)}{2}=\frac{12\times13}{2}=78\). The value 72 may result from incorrectly using 12 in place of \(n+1\). Exam tip: substitute the given value of \(n\) carefully before simplifying the formula.

The terms are (1\cdot2,2\cdot3,3\cdot4,\ldots), so (a_n=n(n+1)). Recognize products of consecutive numbers.

Given (a_n=3^n-2), substitute n=4: (a_4=3^4-2=81-2=79). Therefore, 79 is correct. The closest distractor, 81, is only the value of (3^4); 2 must still be subtracted. Exam tip: for an nth-term formula, substitute the given value of n first, evaluate powers, and then perform the remaining operations.

For odd (n), terms are negative and for even (n), terms are positive, so (a_n=(-1)^n n). Check the sign pattern separately.

To find the first term, put n=1. Thus, a_1=2(1)+9=11, so 11 is correct. The number 9 is only the constant term, not the first term, because the contribution of 2n must also be included when n=1. Exam tip: substitute n=1 in any nth-term formula to obtain the first term.

Given \(a_n=4n^2-1\), substitute \(n=5\) to find the fifth term: \(a_5=4(5)^2-1=4\times25-1=99\). Therefore, 99 is correct. The value 101 would result from incorrectly using \(+1\) instead of the final \(-1\). Exam tip: substitute the term number first and evaluate the exponent before other operations.

The first term is (11) and the difference is (7), so (a_{25}=11+24\cdot7=179). The (25)th term includes (24) differences.

The first five terms are (1,3,5,7,9), and their sum is (25). Listing terms is a good method for small sums.

Given \(a_n=10n+4\), substitute \(n=r+2\): \(a_{r+2}=10(r+2)+4=10r+20+4=10r+24\). Hence, \(10r+24\) is correct. In \(10r+14\), the product \(10\times2=20\) has been handled incorrectly. Exam tip: always substitute the complete index expression in brackets.

The governing pattern is recognition of consecutive cubes and then aligning the cube’s base with the term number. The sequence is 8 = 2³, 27 = 3³, 64 = 4³, and 125 = 5³. The first term uses base 2, so the nth term uses base n + 1. Therefore aₙ = (n + 1)³, making option A correct. Checking the first positions confirms it: at n = 1 the formula gives 2³ = 8, and at n = 4 it gives 5³ = 125. The expression n³ + 1 is not generally equal to (n + 1)³; option C has the wrong type of power, and option D doubles n³ rather than producing consecutive cubes.

(a_2=2) and (a_5=11), so the sum is (13). The listed options do not contain the correct value.

Given \(a_n=2n^2+n\). Substituting \(n=6\), \(a_6=2(6)^2+6=2\times36+6=78\). The option 72 can result from forgetting to add the linear term \(+6\). Exam tip: substitute the value of \(n\) first, then evaluate the exponent carefully.

This is an arithmetic sequence because the common difference between consecutive terms is \(-4\). Here, the first term is \(a=13\) and \(d=-4\). Therefore, \(a_n=a+(n-1)d=13+(n-1)(-4)=17-4n\). Hence, \(17-4n\) is correct. In \(13-4n\), substituting \(n=1\) gives 9 instead of the first term 13. Exam tip: Always substitute \(n=1\) to check whether a proposed nth-term formula gives the first term.

The second differences are (2), and the terms match (n^2+3n). Equal second differences suggest a quadratic rule.

To find when a term is zero, set the explicit rule equal to zero: aₙ = 15 − 2n = 0. Solving gives 2n = 15 and n = 15/2 = 7.5. Sequence indices are positive integers, so there is no integral term number equal to 7.5. Consequently, no term of this sequence is exactly zero, and option A is correct. Checking nearby valid indices confirms this: a₇ = 15 − 14 = 1 and a₈ = 15 − 16 = −1, so the sequence passes from positive to negative without taking zero at an integer index. Options B and C identify neighbouring terms but neither is zero, while D is unrelated.

Given (a_n=n^3-n), substitute n=5 to find the fifth term: (a_5=5^3-5=125-5=120). Hence, the correct answer is 120. The value 125 is only (5^3); the subtraction of 5 must also be included. Exam tip: while finding an nth term, substitute the value of n in every part of the formula.