Summation of series
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| summation/sʌˈmeɪʃn/ | tổng |
| method of differences/ˈmeθəd ɒv ˈdɪfrənsɪz/ | phương pháp hiệu |
| telescope/ˈtelɪskəʊp/ | telescope |
| converge/kənˈvɜːdʒ/ | hội tụ |
| sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ | tổng vô hạn |
| partial fractions/ˈpɑːʃl ˈfrækʃnz/ | phân thức riêng phần |
Count work without adding every stage
- An algorithm performs $r^2$ operations at stage $r$. To estimate its total work after thousands of stages, we need one expression for the whole sum.
- Summation 求和 can combine standard formulas or cancel shifted terms. An infinite total needs a separate check: a finite formula alone does not prove convergence.
Standard sums and their bounds
- For a positive integer $n$, the standard results are:
- The limits matter: $\sum_{r=1}^n1=n$, but $\sum_{r=0}^n1=n+1$. Check the number of terms before applying a formula. The cube identity uses consecutive integers from 1; it is not true for arbitrary lists of numbers.
Summation of series
Sₙ = Σ uₖ
Add the terms one by one — the partial sum climbs toward its limit.
Using Σr = ½n(n+1), what is the sum of the first 5 integers?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
½ × 5 × 6 = 15 (= 1+2+3+4+5).
Using Σr² = (1/6)n(n+1)(2n+1), what is the sum of the first 3 squares?
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(1/6)(3)(4)(7) = 84/6 = 14 (= 1+4+9).
Using Σr³ = ¼n²(n+1)², what is the sum of the first 3 cubes?
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¼ × 9 × 16 = 36 (= 1+8+27). Note: 36 = 6² = (1+2+3)².
Complete the standard result: Σr (from 1 to n) = ______ n(n+1).
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The sum of the first n integers is ½n(n+1).
Expand first, then sum each part
- For $\sum_{r=1}^n(2r+1)(r+2)$, expand to $2r^2+5r+2$. The result is $2\sum r^2+5\sum r+2n$.
- Substitute the standard results and simplify:
At $n=2$, the original terms are $3\times3$ and $5\times4$, totalling 29. The formula also gives $2(16+42+29)/6=29$. A small-$n$ check catches missing constants and index errors.
Σ(2r+1) from r=1 to n = n(n+2). For n=4, find the sum.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
4 × 6 = 24. Check: 3+5+7+9 = 24.
Change the starting or ending index
- If $F(m)=\sum_{r=1}^m u_r$, then $\sum_{r=p}^q u_r=F(q)-F(p-1)$. The last removed term is $p-1$, not $p$.
- For $\sum_{r=0}^n(r+1)^2$, put $k=r+1$. The new bounds are $1$ and $n+1$, so the sum is $(n+1)(n+2)(2n+3)/6$. Change the limits as well as the expression.
In the method of differences, if each term is f(r) − f(r+1), most terms cancel (telescope).
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Consecutive terms cancel, leaving only the first and last pieces.
Σ 1/(r(r+1)) from r=1 to ∞ telescopes to 1 − 1/(n+1). As n→∞, the sum to infinity is?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
As n→∞, 1/(n+1)→0, so the sum approaches 1.
The method of differences 差分法
- Write a term as $F(r)-F(r+1)$, then show the first few terms and the last one. The middle terms telescope 裂项相消, leaving the boundary values $F(1)-F(n+1)$.
- A wider shift leaves more boundary terms. For $F(r)-F(r+2)$, the two starting terms and two ending terms survive:
Partial fractions 部分分式 make cancellation visible
- For $1/[r(r+2)]$, solve $1=A(r+2)+Br$: $A=1/2,B=-1/2$. Thus each term is $\tfrac12(1/r-1/(r+2))$.
- Use the two-step cancellation, keeping its factor:
A decomposition is not yet a sum. Show enough first and last terms to establish the cancellation, especially when three partial fractions or an index shift are involved. Check the formula at $n=1$: it gives $1/3$, the original first term.
What is the sum of the first 4 cubes, and how does it relate to the sum of the first 4 integers?
The cube-sum identity says the sum of the first n cubes is the square of n times (n plus 1) divided by 2.
What is the sum of the first 4 cubes?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The integers sum to 10, and 10 squared is 100. Other options — 36 (the lesson sum of the first 3 cubes); 10 (the sum of the first 4 integers).
Take the limit of the finite sum
- A series converges 收敛 when its partial sums approach a finite real limit. In the last example the two ending fractions tend to zero, so its sum to infinity 无穷和 is $3/4$.
- The remaining tail after $n$ terms is $S_\infty-S_n$. Here it is $\dfrac1{2(n+1)}+\dfrac1{2(n+2)}$, which can be used to choose how many terms give a required accuracy.
Terms tending to zero is not enough. For $u_r=\sqrt{r+1}-\sqrt r$, the terms tend to zero, but $S_n=\sqrt{n+1}-1$ grows without bound. A finite partial-sum limit is the deciding evidence.
Differences need not be partial fractions
- Let $P_r=r(r+1)(r+2)$. Factor the common part to obtain $P_{r+1}-P_r=3(r+1)(r+2)$. Hence a sum of $(r+1)(r+2)$ is $(P_{n+1}-P_1)/3$.
- Another route is to put two fractions over one denominator. For example, $1/r^2-1/(r+1)^2=(2r+1)/[r^2(r+1)^2]$. Recognising the difference avoids trying to use a formula for $\sum1/r^2$.
Check parameter endpoints separately
- Let $u_r=t^{r(r+1)}-t^{(r+1)(r+2)}$. The finite sum is $S_n=t^2-t^{(n+1)(n+2)}$. For $|t|<1$, its limit is $t^2$.
- At $t=1$ or $t=-1$, every exponent is even and every term is zero: the infinite sum is 0, not 1. For $|t|>1$, the ending power grows without bound. The convergence set is $[-1,1]$, but the sum needs separate endpoint values.
Prove the sum before taking infinity
- Choose standard sums after expanding, or differences after rearranging the term. State the index limits, show the surviving boundaries and check a small value of $n$.
- For an infinite series, inspect the limit of the exact partial sum. Check special parameter values in the original terms; use the remaining tail, not just the next term, for an error bound.