Sequences (Mathematics B)

aₙ=a₁+(n−1)d=3+(n−1)·4=4n−1. It is worth checking that n=1 gives a₁=3. The option 4n+3 is the typical error of adding d n times instead of (n−1) times; at n=1 it gives 7, which does not match the first term.

a₁₀=100+(10−1)×(−7)=100−63=37. Adding the common difference 10 times to get 100−70=30 is the typical error; it is added (n−1)=9 times.

The sum of an arithmetic sequence is (first term + last term) × number of terms ÷ 2 = (1+100)×100÷2=5050. Forgetting to divide by 2 gives 10100. The value 4950 is the sum from 1 to 99.

The 10th term is 2+9×3=29, so the sum is 10×(2+29)/2=155. Taking the last term wrongly as 2+10×3=32 gives 10×(2+32)/2=170. The formula Sₙ=n{2a₁+(n−1)d}/2=5×(4+27)=155 gives the same. Forgetting to divide by 2 gives 310, and taking the last term as 9×3=27 without adding the first term gives 10×(2+27)/2=145.

From a₇−a₃=4d=12 we get d=3, and a₁=10−2×3=4, so a₂₀=4+19×3=61. You can also count from a middle term: a₂₀=a₇+13d=22+39=61. The option 64 comes from a₂₀=a₁+20d=4+60, using n instead of (n−1); 58 mistakes the first term for 1; and 67 treats a₃ as the first term and computes 10+19×3.

aₙ=20−3(n−1)=23−3n, and aₙ≧0 is equivalent to n≦7.67, so the terms stay positive up to the 7th, where a₇=2. Then S₇=7×(20+2)/2=77. Once negative terms are added the sum falls, so the maximum is where all the positive terms have been added. The option 70 stops after the 5th term, 80 wrongly takes a₈=0 and computes S₈=8(20+0)/2, and 76 is S₈, which includes the negative term a₈=−1.

The arithmetic mean relation 2x=5+13 gives x=9. The middle term is the average of the two outer ones, (5+13)/2. The value √65 confuses this with the geometric mean, where x²=65.

The multiples of 3 are 3, 6, …, 99, which is 33 terms. The sum is 33×(3+99)/2=33×51=1683. You can also compute 3×(1+2+…+33)=3×561=1683. Do not miscount the number of terms as 32 or 34, since 99÷3=33.

For n≧2 we have aₙ=Sₙ−Sₙ₋₁=(n²+2n)−{(n−1)²+2(n−1)}=2n+1. Also a₁=S₁=3 agrees with 2×1+1=3, so aₙ=2n+1 for every n. The option 2n−1 confuses this with the case Sₙ=n².

a₅=3×2⁵⁻¹=3×2⁴=3×16=48. Using the exponent 5 to get 3×2⁵=96 is the typical error; the ratio is applied (n−1)=4 times.

aₙ=a₁·rⁿ⁻¹=3·2ⁿ⁻¹. Check that n=1 gives a₁=3. The expression 3·2ⁿ gives 6 at n=1, which does not match the first term. Watch out too for 2·3ⁿ⁻¹, which swaps the first term and the common ratio.

S₁₀=(2¹⁰−1)/(2−1)=1024−1=1023. For a common ratio of 2 you can also think of it as twice the last term minus the first term, 2⁹×2−1. The value 1024 is 2¹⁰ itself, and 511 is the sum of the first 9 terms.

S₅=2×(3⁵−1)/(3−1)=2×242/2=242, and indeed 2+6+18+54+162=242. Do not forget the −1 when computing 3⁵−1, and do not forget to divide by the denominator (r−1), which would give 484. The value 243 is 3⁵ itself, and 80 is the sum of the first 4 terms only.

The geometric mean relation x²=2×8=16 gives x=±4. Both the common ratio 2, giving 2, 4, 8, and the common ratio −2, giving 2, −4, 8, work, so the positive value 4 alone is not enough. The value 5 confuses this with the arithmetic mean (2+8)/2.

Substituting n=10 into Σk=n(n+1)/2 gives 10×11/2=55. Forgetting to divide by 2 gives 110. The value 45 is the sum from 1 to 9, so watch the upper limit, and the value 100 comes from computing 10×10.

Substituting n=6 into Σk²=n(n+1)(2n+1)/6 gives 6×7×13/6=91, and indeed 1+4+9+16+25+36=91. The value 55 is the case n=5, and 441=21² confuses this with Σk³=(Σk)². The value 90 does not come from the formula at all; adding the six squares carefully gives 91.

2Σk+Σ1=2×n(n+1)/2+n=n²+n+n=n²+2n. Forgetting to add Σ1=n leaves n²+n. Substituting n=1 to check that you get 3, which is 2×1+1, is worthwhile.

Expand k(k+1)=k²+k, so the sum is Σk²+Σk=385+55=440, using Σk²=10×11×21/6=385 and Σk=55 at n=10. The value 385 stops at Σk² alone, 495 adds twice Σk as 385+110, and 430 uses Σk for n=9, which is 45, giving 385+45. The standard approach is to expand and reduce to the known formulas.

Taking differences gives 1, 2, 3, 4, the natural numbers, so the next difference is 5 and a₆=12+5=17. When the pattern of a sequence is not visible, the standard first move is to examine the differences between neighbouring terms.

The differences are bₖ=k, so for n≧2 we have aₙ=2+Σₖ₌₁ⁿ⁻¹ k=2+(n−1)n/2=(n²−n+4)/2. It also holds at n=1, since 4/2=2. You can check by substituting n=2 and seeing whether you get 3; the expression (n²+n+2)/2 gives 4 at n=2 and is therefore wrong.

Splitting into partial fractions as 1/(k(k+1))=1/k−1/(k+1), the sum becomes (1−1/2)+(1/2−1/3)+…+(1/n−1/(n+1))=1−1/(n+1)=n/(n+1). The middle terms cancel and only the first and the last remain.

Adding 3 each time makes it arithmetic with common difference 3, so a₁₀=1+(10−1)×3=28, and the general term is aₙ=3n−2. Adding 3 ten times to get 31 uses n instead of (n−1).

Multiplying by 3 each time makes it geometric with common ratio 3, so a₄=2×3³=54. Writing the terms out also gives 2, then 6, then 18, then 54. Multiplying by 3 four times to get 162 miscounts the exponent.

Induction has two stages: first, that the statement holds at n=1, and second, that assuming it at n=k lets you show it at n=k+1. Picture knocking over the first domino and showing that whichever one falls knocks over the next; both stages are needed.

The characteristic equation α=2α+1 gives α=−1. Rewriting as aₙ₊₁+1=2(aₙ+1) shows that {aₙ+1} is geometric with first term 2 and common ratio 2, so aₙ+1=2ⁿ and aₙ=2ⁿ−1. This matches the terms 1, 3, 7, 15, 31, …. The option 2ⁿ⁻¹ gives 2 at n=2, which does not match a₂=3.