📚 Summation of Series by the Method of Differences | 差分法求级数求和
The method of differences is a powerful technique for summing finite series whose general term can be expressed as a difference of two consecutive terms of another sequence. If \(u_n = f(n) – f(n-1)\) or \(u_n = f(n+1) – f(n)\), then the sum telescopes, leaving only the first and last terms. This article explains the method step by step, with worked examples and exam-style questions.
差分法是求有限级数求和的一种重要技巧，适用于通项可以表示为另一数列相邻两项之差的级数。如果 \(u_n = f(n) – f(n-1)\) 或 \(u_n = f(n+1) – f(n)\)，则求和时中间项会相互抵消，只剩下首项和末项。本文将逐步讲解该方法，并给出典型例题与考试风格练习。
1. The Core Idea of Telescoping | 裂项相消的核心思想
Consider a sequence \(f(1), f(2), f(3), \dots\). The difference between consecutive terms is \(f(r) – f(r-1)\). If we sum such differences from \(r = 2\) to \(n\), we get:
∑_{r=2}^{n} [f(r) − f(r−1)] = f(n) − f(1)
Similarly, \(\sum_{r=1}^{n} [f(r+1) − f(r)] = f(n+1) − f(1)\). The intermediate terms cancel in pairs, leaving only the endpoints. This is why the method is called “telescoping”.
考虑数列 \(f(1), f(2), f(3), \dots\)。相邻两项之差为 \(f(r) − f(r-1)\)。若对 \(r = 2\) 到 \(n\) 求这些差的和，得到：
∑_{r=2}^{n} [f(r) − f(r−1)] = f(n) − f(1)
同样地，\(\sum_{r=1}^{n} [f(r+1) − f(r)] = f(n+1) − f(1)\)。中间项两两抵消，只剩下端点项，因此这种方法被称为“裂项相消”或“望远镜求和”。
2. Recognising When to Use Differences | 如何判断何时使用差分法
The method works when the general term \(u_r\) can be written as \(g(r) – g(r-1)\) or \(g(r+1) – g(r)\). Common patterns include rational functions whose denominators factor into consecutive integers, products like \(\frac{1}{r(r+1)}\), and trigonometric identities such as \(\sin(r+\theta) – \sin(r-1+\theta)\).
当通项 \(u_r\) 能写成 \(g(r) – g(r-1)\) 或 \(g(r+1) – g(r)\) 时，该方法适用。常见模式包括：分母为相邻整数乘积的有理函数（如 \(\frac{1}{r(r+1)}\)），以及三角恒等式如 \(\sin(r+\theta) − \sin(r−1+\theta)\)。
3. Example: Sum of \(\frac{1}{r(r+1)}\) | 例：求 \(\frac{1}{r(r+1)}\) 的和
We wish to find \(S_n = \sum_{r=1}^{n} \frac{1}{r(r+1)}\). Using partial fractions:
\(\frac{1}{r(r+1)} = \frac{1}{r} − \frac{1}{r+1}\)
Then the sum becomes:
\(S_n = \sum_{r=1}^{n} \left( \frac{1}{r} − \frac{1}{r+1} \right) = 1 − \frac{1}{n+1}\)
For example, \(S_{10} = 1 − \frac{1}{11} = \frac{10}{11}\). As \(n \to \infty\), \(S_n \to 1\).
我们希望求 \(S_n = \sum_{r=1}^{n} \frac{1}{r(r+1)}\)。使用部分分式：
\(\frac{1}{r(r+1)} = \frac{1}{r} − \frac{1}{r+1}\)
于是和式变为：
\(S_n = \sum_{r=1}^{n} \left( \frac{1}{r} − \frac{1}{r+1} \right) = 1 − \frac{1}{n+1}\)
例如，\(S_{10} = 1 − \frac{1}{11} = \frac{10}{11}\)。当 \(n \to \infty\) 时，\(S_n \to 1\)。
4. Generalising the Difference Form | 差分形式的一般化
Sometimes the general term is not immediately a difference, but can be rewritten. For example, \(\frac{1}{r(r+2)}\) can be decomposed as \(\frac{1}{2}\left( \frac{1}{r} − \frac{1}{r+2} \right)\). Writing out the series shows that terms two steps apart cancel, leaving two non-zero terms at each end.
有时通项并不直接是差的形式，但可以改写。例如，\(\frac{1}{r(r+2)}\) 可以分解为 \(\frac{1}{2}\left( \frac{1}{r} − \frac{1}{r+2} \right)\)。写出级数后会看到相隔两项的项相互抵消，两端各留下两个非零项。
5. Worked Example: \(\sum_{r=1}^{n} \frac{1}{r(r+2)}\) | 例题：\(\sum_{r=1}^{n} \frac{1}{r(r+2)}\)
Using the decomposition above:
\(S_n = \frac{1}{2}\sum_{r=1}^{n} \left( \frac{1}{r} − \frac{1}{r+2} \right)\)
Expanding: \( \left(1 − \frac{1}{3}\right) + \left(\frac{1}{2} − \frac{1}{4}\right) + \left(\frac{1}{3} − \frac{1}{5}\right) + \dots + \left(\frac{1}{n} − \frac{1}{n+2}\right) \). The terms \(\frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{n}\) cancel. What remains is:
\(S_n = \frac{1}{2}\left( 1 + \frac{1}{2} − \frac{1}{n+1} − \frac{1}{n+2} \right)\)
For example, \(S_{5} = \frac{1}{2}\left(1.5 − \frac{1}{6} − \frac{1}{7}\right) = \frac{25}{42}\).
利用上述分解：
\(S_n = \frac{1}{2}\sum_{r=1}^{n} \left( \frac{1}{r} − \frac{1}{r+2} \right)\)
展开：\( \left(1 − \frac{1}{3}\right) + \left(\frac{1}{2} − \frac{1}{4}\right) + \left(\frac{1}{3} − \frac{1}{5}\right) + \dots + \left(\frac{1}{n} − \frac{1}{n+2}\right) \)。其中 \(\frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{n}\) 全部抵消，剩下：
\(S_n = \frac{1}{2}\left( 1 + \frac{1}{2} − \frac{1}{n+1} − \frac{1}{n+2} \right)\)
例如，\(S_{5} = \frac{1}{2}\left(1.5 − \frac{1}{6} − \frac{1}{7}\right) = \frac{25}{42}\)。
6. Using Differences for Polynomial Terms | 差分法处理多项式通项
Sometimes we need to sum terms like \(r(r+1)\) or \(r(r+1)(r+2)\). These can be written as differences of higher-degree polynomials. For instance, \(r(r+1) = \frac{1}{3}\left[ (r+1)r(r+2) − r(r−1)(r+1) \right]\). Then the sum telescopes.
有时需要对诸如 \(r(r+1)\) 或 \(r(r+1)(r+2)\) 的项求和。这些可以写成更高次多项式的差。例如，\(r(r+1) = \frac{1}{3}\left[ (r+1)r(r+2) − r(r−1)(r+1) \right]\)，然后求和即可裂项相消。
7. Example: \(\sum_{r=1}^{n} r(r+1)\) | 例：\(\sum_{r=1}^{n} r(r+1)\)
Write \(r(r+1) = \frac{1}{3}\left[ (r+1)r(r+2) − r(r−1)(r+1) \right]\). Summing from \(r=1\) to \(n\):
\(\sum_{r=1}^{n} r(r+1) = \frac{1}{3}\left[ n(n+1)(n+2) − 0 \right] = \frac{n(n+1)(n+2)}{3}\)
Check for \(n=2\): LHS = \(1\cdot2 + 2\cdot3 = 8\), RHS = \(\frac{2\cdot3\cdot4}{3} = 8\). Correct.
将 \(r(r+1) = \frac{1}{3}\left[ (r+1)r(r+2) − r(r−1)(r+1) \right]\)，从 \(r=1\) 到 \(n\) 求和：
\(\sum_{r=1}^{n} r(r+1) = \frac{1}{3}\left[ n(n+1)(n+2) − 0 \right] = \frac{n(n+1)(n+2)}{3}\)
检验 \(n=2\)：左边 = \(1\cdot2 + 2\cdot3 = 8\)，右边 = \(\frac{2\cdot3\cdot4}{3} = 8\)，正确。
8. Trigonometric Differences | 三角差分
Using the product-to-sum identity, \(2\sin A \sin B = \cos(A−B) − \cos(A+B)\). If we let \(u_r = \sin(rx)\), then \(\sin(rx) = \frac{\cos((r−\tfrac{1}{2})x) − \cos((r+\tfrac{1}{2})x)}{2\sin(x/2)}\). Summing over \(r\) gives a telescoping series of cosines.
利用积化和差公式 \(2\sin A \sin B = \cos(A−B) − \cos(A+B)\)。若令 \(u_r = \sin(rx)\)，则 \(\sin(rx) = \frac{\cos((r−\tfrac{1}{2})x) − \cos((r+\tfrac{1}{2})x)}{2\sin(x/2)}\)。对 \(r\) 求和得到一个余弦的裂项相消级数。
9. Example: Sum of Sines | 例：正弦级数求和
Find \(S_n = \sum_{r=1}^{n} \sin(rx)\). Using the identity above:
\(S_n = \frac{1}{2\sin(x/2)} \sum_{r=1}^{n} \left[ \cos\left((r−\tfrac{1}{2})x\right) − \cos\left((r+\tfrac{1}{2})x\right) \right]\)
This telescopes to:
\(S_n = \frac{\cos(x/2) − \cos(n+\tfrac{1}{2})x}{2\sin(x/2)}\)
This result is often used in A-level further mathematics questions.
求 \(S_n = \sum_{r=1}^{n} \sin(rx)\)。利用上述恒等式：
\(S_n = \frac{1}{2\sin(x/2)} \sum_{r=1}^{n} \left[ \cos\left((r−\tfrac{1}{2})x\right) − \cos\left((r+\tfrac{1}{2})x\right) \right]\)
裂项后得到：
\(S_n = \frac{\cos(x/2) − \cos(n+\tfrac{1}{2})x}{2\sin(x/2)}\)
这个结果在A-Level进阶数学中经常出现。
10. Difference of a Product | 乘积型差分
Consider sequences where \(u_r = f(r) – f(r-1)\) and \(f(r)\) itself is a product. For example, \(\frac{1}{r(r+1)(r+2)}\) can be decomposed as:
\(\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left[ \frac{1}{r(r+1)} − \frac{1}{(r+1)(r+2)} \right]\)
Letting \(f(r) = \frac{1}{2r(r+1)}\), we have \(u_r = f(r) − f(r+1)\). Then the sum from \(r=1\) to \(n\) is \(f(1) − f(n+1)\).
考虑 \(u_r = f(r) – f(r-1)\) 且 \(f(r)\) 本身是乘积的形式。例如，\(\frac{1}{r(r+1)(r+2)}\) 可分解为：
\(\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left[ \frac{1}{r(r+1)} − \frac{1}{(r+1)(r+2)} \right]\)
令 \(f(r) = \frac{1}{2r(r+1)}\)，则 \(u_r = f(r) − f(r+1)\)。因此从 \(r=1\) 到 \(n\) 的和为 \(f(1) − f(n+1)\)。
11. Working with \(\frac{r}{(r+1)!}\) | 处理 \(\frac{r}{(r+1)!}\)
Factorials also allow telescoping. Noticing that \(\frac{r}{(r+1)!} = \frac{r+1−1}{(r+1)!} = \frac{1}{r!} − \frac{1}{(r+1)!}\). Hence:
\(\sum_{r=1}^{n} \frac{r}{(r+1)!} = 1 − \frac{1}{(n+1)!}\)
This is a classic example showing how algebraic manipulation reveals a hidden difference.
阶乘同样可以裂项。注意到 \(\frac{r}{(r+1)!} = \frac{r+1−1}{(r+1)!} = \frac{1}{r!} − \frac{1}{(r+1)!}\)。因此：
\(\sum_{r=1}^{n} \frac{r}{(r+1)!} = 1 − \frac{1}{(n+1)!}\)
这是一个经典例子，说明代数变形如何揭示隐藏的差的形式。
12. Exam Tips and Common Pitfalls | 考试技巧与常见错误
1. Always write out the first few terms to see which terms actually cancel.
2. Be careful with coefficients from partial fractions – forgetting the factor \(\frac{1}{2}\) or \(\frac{1}{3}\) is a common error.
3. When summing from \(r=1\) to \(n\), the index in the difference may start at different values; ensure the limits are correct.
4. For infinite series, take the limit \(n \to \infty\) only if the remainder tends to zero.
- 1. 务必写出前几项，观察哪些项会真正抵消。
- 2. 注意部分分式产生的系数，漏写 \(\frac{1}{2}\) 或 \(\frac{1}{3}\) 是常见错误。
- 3. 求和范围从 \(r=1\) 到 \(n\) 时，差式中的起始下标可能不同，务必确认上下限正确。
- 4. 对无穷级数，只有当剩余项趋于零时，才能取极限 \(n \to \infty\)。
By mastering the method of differences, you can solve a wide variety of summation problems quickly and accurately. Practice rewriting each general term until the difference structure becomes obvious.
掌握差分法后，你可以快速准确地解决多种求和问题。练习时将每个通项改写为差的形式，直到差分结构变得明显。
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