📚 Summation of Series by the Method of Differences | 差分法求级数和
The method of differences is a powerful AQA A-Level Mathematics tool for summing finite series. It is especially useful when a general term ur can be expressed as f(r) − f(r+1) or f(r+1) − f(r). The resulting cancellation collapses the sum to a simple closed form, and this single idea underpins almost every telescoping question on the exam paper.
差分法是 AQA A-Level 数学中求有限级数和的重要方法。当通项 ur 可以写成 f(r) − f(r+1) 或 f(r+1) − f(r) 时，求和过程中大部分项会消去，最终得到一个简洁的闭合形式。正是这一个思路，支撑了试卷上几乎所有"望远镜求和"题目。
1. What is the Method of Differences? | 什么是差分法？
The method of differences applies when the r-th term of a series can be expressed as the difference between two consecutive terms of another sequence. Instead of adding n terms directly, we add these differences and observe that most intermediate values cancel in pairs. The method is also called telescoping summation, because the surviving terms look like the two ends of a collapsed telescope.
差分法适用于通项可以写成另一个序列中相邻两项之差的情形。我们不需要直接相加 n 项，而是将这些差分相加，并发现大多数中间项成对抵消。这个方法也叫望远镜求和，因为最后留下来的项就像被压缩的望远镜的两端。
to use it, or you must produce the split yourself using partial fractions. In both cases the final answer is a closed form in n, not a long decimal expansion.
在 AQA 考试中，这一技巧通常以两种形式出现：要么题目直接给出恒等式（例如 1/(r(r+1)) = 1/r − 1/(r+1)）让你使用；要么需要你自己用部分分式完成拆分。无论哪种形式，最终答案都是关于 n 的闭合表达式，而不是一长串小数。
2. The Two Fundamental Identities | 两个基本恒等式
Everything in this topic follows from two standard results. The first deals with the form f(r) − f(r+1), where each term cancels with the next term's first half.
本专题的一切都源于两个标准结论。第一个针对 f(r) − f(r+1) 这种形式，其中每一项与下一项的前半部分相消。
∑r=1n [f(r) − f(r+1)] = f(1) − f(n+1)
Only the very first piece f(1) and the very last piece −f(n+1) survive; every f(2), f(3), … , f(n) appears twice with opposite signs.
只有最前面的 f(1) 和最后面的 −f(n+1) 保留下来；每一个 f(2)、f(3)、…、f(n) 都出现两次且符号相反。
second identity handles the reversed order, where each term cancels with the previous term' n becomes f(n) − f(0). The key idea is exactly the same: only the first and last parts survive.
第二个恒等式处理相反的顺序，即每一项与上一项的后半部分相消。如果把项写成 f(r) − f(r−1)，则从 r = 1 到 n 的和为 f(n) − f(0)。核心思想完全相同：只有首尾部分保留下来。
∑r=1n [f(r) − f(r−1)] = f(n) − f(0)
A useful equivalent version, often easier to apply, is ∑r=1n r+1) − f(r)] = f(n+1) − f(1). Choose whichever orientation makes the cancellation visible.
一个同样常用、往往更好用的等价形式是 ∑r=1n [f(r+1) − f(r)] = f(n+1) − f(1)。选择哪种方向，取决于哪种写法能让消项过程更清楚。
3. Why Cancellation Happens: The Telescoping Picture | 为什么会消项：望远镜图景
n = 4, using the first identity. The expansion makes the mechanism completely transparent.
用第一个恒等式把 n = 4 的情形展开，消项的机制就一目了然。
[f(1) − f(2)] + [f(2) − f(3)] + [f(3) − f(4)] + [f(4) − f(5)]
f(4) cancels the +f(4). What remains is f(1) − f(5), which matches the formula with n = 4.
第一个括号中的 −f(2) 与第二个括号中的 +f(2) 相消；−f(3) 与 +f(3) 相消；−f(4) 与 +f(4) 相消。最后剩下 f(1) − f(5)，与 n = 4 时的公式完全一致。
This is why examiners insist that you show the cancellation explicitly for the general term, not merely for a numerical list. A safe habit is to write the first two terms, an ellipsis, and the last two terms, so the marker can see exactly which pieces disappear.
这也是为什么考官要求你必须对一般项展示消项过程，而不能只列几个数字项。一个稳妥的习惯是写出前两项、省略号、以及最后两项，让阅卷人清楚看到哪些部分消失了。
4. The Standard Five-Step Recipe | 标准五步法
Most method-of-differences questions can be solved by following the same five steps in order. Practising this routine until it is automatic saves a great deal of time in the exam.
大多数差分法题目都可以按同样的五步依次解决。把这套流程练到自动化，能在考试中节省大量时间。
- Step 1 – Inspect the general term: identify ur and decide whether it is a rational, factorial or trigonometric expression.
第一步——观察通项：确认 ur，判断它是分式型、阶乘型还是三角型表达式。 - Step 2 – Split it: use partial fractions, factorial manipulation or a given identity to write ur as f(r) − f(r+1).
第二步——拆分：用部分分式、阶乘变形或题目给出的恒等式，把 ur 写成 f(r) − f(r+1)。 - Step 3 – Expand: write out the first two and last two terms to expose the cancellation pattern.
第三步——展开：写出前两项和后两项，暴露消项规律。 - Step 4 – Collapse: apply the identity to obtain f(1) − f(n+1).
第四步——合并：套用恒等式得到 f(1) − f(n+1)。 - Step 5 – Simplify: combine into a single fraction and, if asked, take the limit as n → ∞.
第五步——化简：合并成一个分式，若题目要求，再求 n → ∞ 时的极限。
Step 2 is where most marks are lost, because an incorrect split produces an answer that cannot cancel. Always verify your split by recombining the fractions over a common denominator.
第二步最容易丢分，因为拆分错误会导致无法消项。务必把拆开的分式通分回原来的形式，以检验拆分是否正确。
5. Splitting the General Term: Partial Fractions | 拆分通项：部分分式
The most common splits in AQA questions involve two or three linear factors. Learn these three patterns by heart: they cover the overwhelming majority of examination items.
AQA 题目中最常见的拆分涉及两个或三个一次因式。把下面三种模式背熟，就能覆盖考试中绝大多数题目。
1/(r(r+1)) = 1/r − 1/(r+1)
1/(r(r+k)) = (1/k)[1/r − 1/(r+k)] , k ≠ 0
1/(r(r+1)(r+2)) = ½[ 1/(r(r+1)) − 1/((r+1)(r+2)) ]
The third pattern is a two-level telescope: the split is not into simple fractions but into a difference of the two-factor expression, which then cancels one level down. Notice that the ½ is essential and must not be dropped.
第三种模式是"两级望远镜"：它不是拆成最简分式，而是拆成两个二因式表达式之差，然后再向下一层消项。注意其中的 ½ 必不可少，绝不能漏掉。
= 1/r² − 1/(r+1)², a result worth remembering because it appears regularly in past papers.
当分子不是 1 时，思路相同，但要特别小心系数。例如 (2r+1)/(r²(r+1)²) = 1/r² − 1/(r+1)²，这个结果值得记住，因为它在历年真题中反复出现。
The same philosophy extends beyond algebra: factorial terms such as r·r! can be written as (r+1)! − r!, and certain trigonometric products split using the addition formulae. The underlying identity of the method never changes.
同样的思想不限于代数：阶乘型通项如 r·r! 可以写成 (r+1)! − r!；某些三角乘积也可以借助和角公式拆分。差分法的核心恒等式始终不变。
6. Worked Example 1: The Classic 1/(r(r+1)) | 例题一：经典情形
Find the sum of the series ∑r=1n 1/(r(r+1)) and deduce its sum to infinity.
求级数 ∑r=1n 1/(r(r+1)) 的和，并由此推出它的无穷和。
itting into partial fractions gives 1/(r(r+1)) = 1/r − 1/(r+1), so we may take f(r) = 1/r.
第一步：通项为 1/(r(r+1))。第二步：部分分式拆分得 1/(r(r+1)) = 1/r − 1/(r+1)，因此可取 f(r) = 1/r。
Step 3 and 4: expanding and cancelling, the intermediate terms 1/2, 1/3, … , 1/n all disappear in pairs, leaving the first term 1/1 and the final term −1/(n+1).
第三、四步：展开并消项后，中间项 1/2、1/3、…、1/n 都会成对消失，只剩下首项 1/1 与末项 −1/(n+1)。
Sn = 1 − 1/(n+1) = n/(n+1)
e harmonic series ∑ 1/r diverges, which is a point worth commenting on in a longer answer.
第五步：当 n → ∞ 时，1/(n+1) → 0，所以无穷和为 1。虽然调和级数 ∑ 1/r 发散，但这个级数却是收敛的，在较长的解答中值得点明这一点。
7. Worked Example 2: Three Factors in the Denominator | 例题二：分母含三个因式
Find ∑r=1n 1/(r(r+1)(r+2)) and state the sum to infinity.
求 ∑r=1n 1/(r(r+1)(r+2))，并写出无穷和。
n, the general term becomes ½[1/(r(r+1)) − 1/((r+1)(r+2))], so here f(r) = ½ · 1/(r(r+1)).
利用三因式模式，通项可写成 ½[1/(r(r+1)) − 1/((r+1)(r+2))]，此时 f(r) = ½ · 1/(r(r+1))。
Expanding for r = 1, 2, 3, … s: the −1/((r+1)(r+2)) at r = 1 matches the +1/(r(r+1)) at r = 2, and so on down the line.
把 r = 1, 2, 3, … , n 展开，除首尾外的每一块都相
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