Sequences

A quadratic sequence usually has the form \(an^2+bn+c\). Its first differences change, but the second difference is constant and equals \(2a\). An arithmetic sequence has a constant first difference instead. Exam tip: make a difference table to identify it quickly.

The terms are of the form (n^2+1), so the next term is (6^2+1=37). Exam tip: recognize patterns built around squares.

The formula \(a_n=n(n+1)\) says that the \(n\)-th term is the product of two consecutive positive integers, \(n\) and \(n+1\). To find which term equals 210, we need to solve \(n(n+1)=210\). We can look for two consecutive factors of 210. The pair 14 and 15 works because both are consecutive and their product is 210.

Thus \(n=14\), so 210 is the 14th term and option A is correct. Substitution verifies it: \(a_{14}=14(14+1)=14\times15=210\). The 15th term would instead be \(15\times16=240\), so option B is not correct. The index is the first factor \(n\), not the second factor \(n+1\).

The direct answer is option C: 231. Look at the differences between consecutive terms: 6-3=3, 10-6=4, 15-10=5, and 21-15=6. The differences increase by 1 each time. This is the triangular-number pattern, and its nth term is \\(a_n=\\frac{(n+1)(n+2)}{2}\\). For n=20, substitute carefully: \\(a_{20}=\\frac{(20+1)(20+2)}{2}=\\frac{21×22}{2}=21×11=231\\). Therefore option C is correct. Option A, 210, equals \\(20×21/2\\), which is the triangular number beginning with 1 and does not match this sequence. Option B, 220, does not result from the given formula or pattern. Option C, 231, is the calculated twentieth term. Option D, 242, is also not produced by the formula and is too large. Another way to see the indexing is that the first term is \\(2×3/2=3\\), the second is \\(3×4/2=6\\), and the fifth is \\(6×7/2=21\\). The common mistake is to use \\(n(n+1)/2\\) without noticing that this sequence starts one step later.

The successive differences \(4,9,16,25\) are \(2^2,3^2,4^2,5^2\). Hence, the next difference is \(6^2=36\). Therefore, the next term is \(55+36=91\). Taking a difference such as \(30\) after \(25\) would be incorrect because the differences follow consecutive perfect squares. Exam tip: when a sequence pattern is unclear, list its successive differences first.

The differences between consecutive terms are \(11-2=9\), \(26-11=15\), \(47-26=21\), and \(74-47=27\). These differences increase by \(6\) each time, so the next difference is \(33\). Therefore, the next term is \(74+33=107\). Choosing \(105\) would give a difference of \(31\), which does not follow the pattern. Exam tip: For a sequence, first list consecutive differences and look for their pattern.

Odd positions are (3,6,12,\ldots), so the next odd-position term is (24). Exam tip: split alternating sequences by positions.

After (34), multiplication by (2) is applied, so the next term is (68). Exam tip: identify the next operation in alternating-operation sequences.

Apply the recurrence step by step: \(a_2=2(1)+1=3\), \(a_3=2(3)+2=8\), \(a_4=2(8)+3=19\), and \(a_5=2(19)+4=42\). Therefore, the correct answer is \(42\). A choice such as \(44\) can result from using an incorrect value of \(n\) in the final step. Exam tip: write the value of \(n\) for each step before substituting it in a recurrence relation.

Substituting n=1,2,3,4,5,6 successively in the recurrence gives a_2=7, a_3=12, a_4=19, a_5=28, a_6=39, and a_7=52. Hence, the correct answer is 52. A value such as 50 can result from adding an incorrect increment instead of 2n+1 at one step. Exam tip: for each new term, use the index of the preceding term as n.

Given \(a_n=(-1)^{n+1}n^2\), \(a_9=(-1)^{10}\times9^2=81\) and \(a_{10}=(-1)^{11}\times10^2=-100\). Hence, \(a_9+a_{10}=81-100=-19\). The option \(-181\) may result from adding both magnitudes and then assigning a negative sign, which is incorrect. Exam tip: check the sign of \((-1)^{n+1}\) separately for odd and even values of \(n\).

The fifth term is 121. Using the given rule, the sixth term is \(3\times121+1=363+1=364\). Option 365 would result from incorrectly adding 2 instead of 1. Exam tip: in a recursive sequence, apply the stated rule to the immediately preceding term.

Given \(a_n=2^n+n\), substitute \(n=6\): \(a_6=2^6+6=64+6=70\). Therefore, option B is correct. The value 68 would result from adding 4 to \(2^6\), which does not follow the given nth-term rule. Exam tip: while finding a term of a sequence, substitute the value of \(n\) in every required place and evaluate the power first.

Each term in this sequence is obtained by doubling the previous term and adding 1. Its nth term is \(a_n=2^n-1\). Since \(2^7-1=128-1=127\), 127 is the seventh term. The sixth term is \(2^6-1=63\), so it is not correct. Exam tip: add 1 to the terms to identify powers of 2 in such sequences.

This is (2^2,3^2,4^2,\ldots), so (169=13^2) is the (12)th term. Exam tip: count positions carefully in shifted square sequences.

The consecutive differences are 5, 9, 13, and 17. Since the difference after 12 is 9, x = 12 + 9 = 21. Check: 21 + 13 = 34 and 34 + 17 = 51, so the sequence is consistent. If 20 were chosen, the next difference would be 14, not the given 13. Exam tip: for a missing term in a sequence, verify the differences on both sides of the term.

The consecutive differences are \(-4,-7,-10,-13\). Therefore, adding \(-10\) to the third term gives \(x=89-10=79\). Check: \(79+(-13)=66\), so the next difference is also satisfied. If 76 were chosen, the difference from 89 would be \(-13\), not the required third difference of \(-10\). Exam tip: a negative difference means subtract that value from the preceding term.

This is a geometric sequence with first term \(a=2\) and common ratio \(r=3\). Its \(n\)th term is \(a_n=ar^{n-1}\). Therefore, \(a_8=2\times3^{8-1}=2\times3^7=4374\). Option 2187 is \(3^7\), but it misses multiplication by the first term, 2. Exam tip: for a geometric sequence, use exponent \(n-1\) for the \(n\)th term.

The terms are (512,256,128,64,32,16,8), so the (7)th term is the first smaller term. Exam tip: move stepwise until the condition is met.

In \(1,4,9,16,\ldots\), the consecutive differences are \(3,5,7,\ldots\), so they are not constant; the ratios are not constant either. Hence it is neither AP nor GP. Exam tip: check both differences and ratios.

To identify a general term, substitute small positive values of \(n\) into each proposed formula and compare them with the listed sequence. For the proposed expression \(a_n=2n^2+n+2\), at \(n=1\) we get \(2+1+2=5\); at \(n=2\), we get \(8+2+2=12\); and at \(n=3\), we get \(18+3+2=23\). These match the first three terms.

The formula also gives \(a_4=2(4^2)+4+2=32+4+2=38\) and \(a_5=2(5^2)+5+2=50+5+2=57\), matching the next terms as well. Hence option D is correct. Checking several terms is important because matching only one term could happen by chance. The other formulas fail when tested against the sequence.

For option A, the first differences are \(4,6,8\), so the second differences are \(2,2\), which are equal and non-zero. In B, the second difference is \(0\). Exam tip: list two levels of differences carefully.

(a_{12}=192) and (a_{13}=221), so (12) terms are less than (200). Exam tip: check the two terms near the boundary.

The question asks for the sum of the first five terms, so the listed terms from the first through the fifth must all be included exactly once. They are 2, 5, 10, 17, and 26. This is not asking for the fifth term alone. One may notice that the terms follow the rule \(n^2+1\), but direct addition is the simplest method for the requested small number of terms.

Add the terms in order: \(2+5=7\), \(7+10=17\), \(17+17=34\), and \(34+26=60\). Therefore the sum is 60, so option D is correct. Omitting the first term would give an incorrect total, and stopping at the fourth term would give only 34. Every one of the five specified terms must be counted.

Successive differences mean that each new term is obtained by adding the stated difference to the preceding term. Thus, \(a_2=2+3\), \(a_3=a_2+7\), \(a_4=a_3+11\), and \(a_5=a_4+15\). Therefore, \(a_5=2+3+7+11+15=38\). Option 40 would result from incorrectly adding 2 extra to the total. Exam tip: to find \(a_n\), add the first \(n-1\) successive differences to \(a_1\).