Series
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| geometric series/ˌdʒiːəʊˈmetrɪk ˈsɪəriːz/ | cấp số nhân |
| converges/kənˈvɜːdʒɪz/ | hội tụ |
| binomial expansion/baɪˈnəʊmɪəl ekˈspænʃn/ | nhị thức Newton |
| binomial coefficient/baɪˈnəʊmɪəl ˌkəʊɪˈfɪʃənt/ | hệ nhị thức |
| arithmetic progression/əˈrɪθmətɪk prəˈɡreʃn/ | cấp số cộng |
| common difference/ˈkɒmən ˈdɪfrəns/ | công sai |
| geometric progression/ˌdʒiːəʊˈmetrɪk prəˈɡreʃn/ | cấp số nhân |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | công bội |
| sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ | tổng vô hạn |
Can an endless payment have a finite total?
- Imagine payments of $8,-4,2,-1,\ldots$: a payment followed by a smaller refund. The running total moves above and below one value.
- This geometric series 等比级数 converges 收敛: its partial sums approach a finite limit. We will find that limit and distinguish a term from a sum.
Build a binomial expansion 二项展开式
- For a positive integer $n$, choose the second part $k$ times and the first part $n-k$ times. The corresponding term is $\binom nk a^{n-k}b^k$, for $k=0,1,\ldots,n$.
- The binomial coefficient 二项式系数 is $\binom nk=\dfrac{n!}{k!(n-k)!}$, with $0!=1$. Include the sign and the whole power of each part.
For $(1-2x)^4$, the five terms are $1-8x+24x^2-32x^3+16x^4$. The alternating signs come from $(-2x)^k$, not from the coefficients $1,4,6,4,1$.
What is the binomial coefficient C(5, 2) = 5! / (2! 3!)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
C(5,2) = 120/(2×6) = 120/12 = 10.
Find one coefficient without expanding everything
- In $(x^2+2/x)^6$, term $k$ is $\binom6k2^k x^{12-3k}$. Set the exponent to zero for the constant term: $k=4$, giving $15\times16=240$.
- In a product, list every pair of powers that makes the required power. For the coefficient of $x^2$ in $(1+3x)(2-x)^4$, use $1\times24x^2$ and $3x\times(-32x)$: the coefficient is $24-96=-72$.
A term includes its power of $x$; a coefficient is the number multiplying that power. A negative power of $x$ can contribute to a positive power after multiplication. Check every possible pairing.
Expand $(x-3/x^2)^4$ completely. Separately find the constant term in $(2x^2-1/x)^6$.
The expansion is $x^4-12x+54x^{-2}-108x^{-5}+81x^{-8}$. In the second expression term k has power $12-3k$; k=4 gives $\binom64 2^2(-1)^4=60$.
The coefficient of $x^2$ in $(2-px)^5(1+3x)$ is −160. The coefficient of x is positive. Find p, showing both constraints.
The first terms of $(2-px)^5$ are $32-80px+80p^2x^2$. The product has x² coefficient $80p^2-240p=-160$, so $(p-1)(p-2)=0$. Its x coefficient is $96-80p>0$, which selects p=1 and rejects p=2.
Arithmetic progressions 等差数列
- An arithmetic progression (AP) adds a fixed common difference 公差 $d$. From first term $a$, the $n$th term and sum of the first $n$ terms are:
- For three consecutive terms $a,b,c$, equal differences give $2b=a+c$. To find unknown first term and difference, write two term equations before solving them.
If $u_3=13$ and $u_8=33$, then $a+2d=13$ and $a+7d=33$. Subtract: $5d=20$, so $d=4$ and $a=5$. Hence $u_{10}=41$ and $S_{10}=5(5+41)=230$.
An AP has first term a = 3 and common difference d = 4. What is the 5th term?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
u₅ = a + (n−1)d = 3 + 4×4 = 3 + 16 = 19.
An AP has a = 2 and d = 3. What is the sum of the first 5 terms?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
S₅ = (5/2)(2×2 + 4×3) = (5/2)(4 + 12) = (5/2)(16) = 40.
An AP has fourth term 11 and tenth term −7. Find a and d, its eighteenth term, the sum of its first 18 terms, and the number of positive terms.
$a+3d=11$, $a+9d=-7$ give d=−3,a=20. Then $u_{18}=20-51=-31$ and $S_{18}=9(20-31)=-99$. $u_n=23-3n>0$ gives n<23/3, so there are 7 positive terms.
Choose the right part of an AP
- “Terms between two values” means choose their positions first. For $a=4,d=3$, terms strictly between $20$ and $41$ satisfy $20<3n+1<41$, so $n=7,\ldots,13$.
- There are $13-7+1=7$ terms, from $22$ to $40$. Their sum is $\dfrac72(22+40)=217$, also $S_{13}-S_6$. Do not count the sixth term again.
An AP starts at 5 and has common difference 7. Find the sum of all its terms strictly between 40 and 110. State the first and last included positions.
$u_n=7n-2$. $40<7n-2<110$ gives $6<n<16$, so positions 7 through 15 are included. The terms are 47 through 103: 9 terms, sum $9(47+103)/2=675$.
Geometric progressions 等比数列
- A geometric progression (GP) multiplies by a fixed common ratio 公比 $r$. For non-zero consecutive terms, divide the later by the earlier; for three consecutive terms $a,b,c$, $b^2=ac$.
- If $r=1$, every term is $a$ and $S_n=na$. A constant non-zero sequence is both an AP ($d=0$) and a GP ($r=1$). A negative ratio makes signs alternate.
Build a sequence
uₙ = a · rⁿ⁻¹
Each term multiplies by r. When |r| < 1 the terms shrink and the sum to infinity converges.
A GP has a = 2 and r = 3. What is the 4th term?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
u₄ = ar³ = 2 × 3³ = 2 × 27 = 54.
Classify these as AP, GP, both or neither, giving d or r where possible: (i) 7,4,1,−2; (ii) 8,−4,2,−1; (iii) 4,4,4,4; (iv) 1,2,4,7. Which have a finite sum to infinity if their rules continue?
(i) AP, d=−3. (ii) GP, r=−1/2, finite sum 16/3. (iii) Both, d=0 and r=1, no finite infinite sum. (iv) Neither: neither differences nor ratios are constant; the listed four terms alone do not specify its continuation.
Connect an AP to a GP
- Keep one meaning for each letter. Suppose $p,q,s$ are consecutive GP terms, while $p,q,-2s$ are consecutive AP terms. Then $q^2=ps$ and $2q=p-2s$.
- Put $q=pr,s=pr^2$ for $p\ne0$. The AP equation becomes $2r=1-2r^2$. Solve $2r^2+2r-1=0$, then apply any sign or convergence condition in the question.
An AP condition and a GP condition are two separate equations. Do not use a common difference in the GP equation or assume both progressions have the same third term.
Sum to infinity 无穷和 and the remaining tail
- For a non-zero first term, the series converges exactly when $|r|<1$. Since $r^n\to0$, the finite-sum formula gives $S_\infty=\dfrac a{1-r}$. Terms tending to zero alone do not establish convergence for every kind of series.
- After removing the first $n$ terms, the remaining sum starts with $ar^n$: $S_\infty-S_n=\dfrac{ar^n}{1-r}$. It can be negative; its magnitude is the error in using $S_n$.
If the second term is $-12$ and $S_\infty=16$, then $ar=-12$ and $a=16(1-r)$. Thus $4r^2-4r-3=0$, giving $r=-1/2$ or $3/2$. Convergence rejects $3/2$, so $a=24$. After three terms the remaining sum is $24(-1/2)^3/(3/2)=-2$.
A GP has a = 8 and r = 0.5. What is the sum to infinity?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
S∞ = a/(1−r) = 8/(1−0.5) = 8/0.5 = 16.
A geometric series with a non-zero first term and r=1.5 has a finite sum to infinity.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The sum to infinity formula only applies when |r| < 1. With r = 1.5, the series diverges.
Match each formula to what it computes.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
d marks the arithmetic family, r the geometric one — recognising the formula is half the exam question.
A GP has second term −36 and sum to infinity 48. Find a and r, showing why one algebraic root is rejected. Find S₄ and the sum after removing its first four terms.
$ar=-36$, $a=48(1-r)$ give $4r^2-4r-3=0$. The roots are −1/2 and 3/2; reject 3/2 since |r|<1 is required. Thus a=72, $S_4=72-36+18-9=45$, and the remaining sum is 48−45=3.
The positive numbers a,b,c are consecutive GP terms. The numbers a,2b,−3c are consecutive AP terms. Find the GP ratio exactly. If a=6, find its sum to infinity.
Write b=ar,c=ar², with r>0. The AP gives $4b=a-3c$, hence $3r^2+4r-1=0$. The positive root is $r=(\sqrt7-2)/3$, which lies between 0 and 1. Then $S_\infty=6/(1-r)=5+\sqrt7$.
Choose, calculate, check
- Identify whether the question asks for a term, a finite sum, a selected range or an infinite tail. State $|r|<1$ before using an infinite-sum formula; for $r=-1$, partial sums oscillate without settling.
- For a binomial term, track both its coefficient and its exponent. For progressions, check your answer against the original term equations and any positive, negative or integer conditions.
An arithmetic series has first term 2, common difference 5 and 6 terms. What are the sixth term and the sum?
Five steps of 5 from 2 reach 27. Three times 29 is 87.
An AP has first term 2, difference 5 and 6 terms. What is the sum of those 6 terms?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
6/2 times 29 is 87.
A GP starts at 12 with ratio −1/3. Find the least positive integer n for which the magnitude of the remaining infinite sum after n terms is less than 0.01. Give Sₙ for that n.
$S_\infty=9$ and the remaining sum is $9(-1/3)^n$. Its magnitude is $9/3^n$. At n=6 it is 1/81>0.01; at n=7 it is 1/243<0.01. Since these magnitudes decrease, the least n is 7. $S_7=9-9(-1/3)^7=9+1/243$.