Summation of Series by the Method of Differences | 差分法求级数求和

📚 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。

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1. The Core Idea of Telescoping | 裂项相消的核心思想

Consider a sequence f(1), f(2), f(3), …. The difference between consecutive terms is f(r) - f(r-1). If we sum such differences from r = 2 to n, we get:

∑_r=2ⁿ [f(r) − f(r−1)] = f(n) − f(1)

Similarly, ∑_r=1ⁿ [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), …。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=2ⁿ [f(r) − f(r−1)] = f(n) − f(1)

同样地，∑_r=1ⁿ [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”。

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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 1r(r+1), and trigonometric identities such as sin(r+θ) - sin(r-1+θ).

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 1r(r+1)），以及三角恒等式如 sin(r+θ) − sin(r−1+θ)。

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3. Example: Sum of 1r(r+1) | 例：求 1r(r+1) 的和

We wish to find S_n = ∑_r=1ⁿ 1r(r+1). Using partial fractions:

1r(r+1) = 1r − 1r+1

Then the sum becomes:

S_n = ∑_r=1ⁿ ( 1r − 1r+1 ) = 1 − 1n+1

For example, S₁₀ = 1 − 111 = 1011. As n → ∞, S_n → 1.

我们希望求 S_n = ∑_r=1ⁿ 1r(r+1)。使用部分分式：

1r(r+1) = 1r − 1r+1

于是和式变为：

S_n = ∑_r=1ⁿ ( 1r − 1r+1 ) = 1 − 1n+1

例如，S₁₀ = 1 − 111 = 1011。当 n → ∞ 时，S_n → 1。

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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, 1r(r+2) can be decomposed as 12( 1r − 1r+2 ). 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，1r(r+2) 可以分解为 12( 1r − 1r+2 )。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。

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5. Worked Example: ∑_r=1ⁿ 1r(r+2) | 例题：∑_r=1ⁿ 1r(r+2)

Using the decomposition above:

S_n = 12∑_r=1ⁿ ( 1r − 1r+2 )

Expanding: (1 − 13) + (12 − 14) + (13 − 15) + … + (1n − 1n+2) . The terms 13, 14, …, 1n cancel. What remains is:

S_n = 12( 1 + 12 − 1n+1 − 1n+2 )

For example, S₅ = 12(1.5 − 16 − 17) = 2542.

利用上述分解：

S_n = 12∑_r=1ⁿ ( 1r − 1r+2 )

展开： (1 − 13) + (12 − 14) + (13 − 15) + … + (1n − 1n+2) 。其中 13, 14, …, 1n 全部抵消，剩下：

S_n = 12( 1 + 12 − 1n+1 − 1n+2 )

例如，S₅ = 12(1.5 − 16 − 17) = 2542。

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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) = 13[ (r+1)r(r+2) − r(r−1)(r+1) ]. 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) = 13[ (r+1)r(r+2) − r(r−1)(r+1) ]，然后求和即可裂项相消。

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7. Example: ∑_r=1ⁿ r(r+1) | 例：∑_r=1ⁿ r(r+1)

Write r(r+1) = 13[ (r+1)r(r+2) − r(r−1)(r+1) ]. Summing from r=1 to n:

∑_r=1ⁿ r(r+1) = 13[ n(n+1)(n+2) − 0 ] = n(n+1)(n+2)3

Check for n=2: LHS = 1·2 + 2·3 = 8, RHS = 2·3·43 = 8. Correct.

将 r(r+1) = 13[ (r+1)r(r+2) − r(r−1)(r+1) ]，从 r=1 到 n 求和：

∑_r=1ⁿ r(r+1) = 13[ n(n+1)(n+2) − 0 ] = n(n+1)(n+2)3

检验 n=2：左边 = 1·2 + 2·3 = 8，右边 = 2·3·43 = 8，正确。

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8. Trigonometric Differences | 三角差分

Using the product-to-sum identity, 2sin A sin B = cos(A−B) − cos(A+B). If we let u_r = sin(rx), then sin(rx) = ^{cos((r−12)x) − cos((r+12)x)}⁄_2sin(x/2). Summing over r gives a telescoping series of cosines.

利用积化和差公式 2sin A sin B = cos(A−B) − cos(A+B)。若令 u_r = sin(rx)，则 sin(rx) = ^{cos((r−12)x) − cos((r+12)x)}⁄_2sin(x/2)。对 r 求和得到一个余弦的裂项相消级数。

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9. Example: Sum of Sines | 例：正弦级数求和

Find S_n = ∑_r=1ⁿ sin(rx). Using the identity above:

S_n = 12sin(x/2) ∑_r=1ⁿ [ cos((r−12)x) − cos((r+12)x) ]

This telescopes to:

S_n = ^{cos(x/2) − cos(n+12)x}⁄_2sin(x/2)

This result is often used in A-level further mathematics questions.

求 S_n = ∑_r=1ⁿ sin(rx)。利用上述恒等式：

S_n = 12sin(x/2) ∑_r=1ⁿ [ cos((r−12)x) − cos((r+12)x) ]

裂项后得到：

S_n = ^{cos(x/2) − cos(n+12)x}⁄_2sin(x/2)

这个结果在A-Level进阶数学中经常出现。

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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, 1r(r+1)(r+2) can be decomposed as:

1r(r+1)(r+2) = 12[ 1r(r+1) − 1(r+1)(r+2) ]

Letting f(r) = 12r(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，1r(r+1)(r+2) 可分解为：

1r(r+1)(r+2) = 12[ 1r(r+1) − 1(r+1)(r+2) ]

令 f(r) = 12r(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)。

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11. Working with r(r+1)! | 处理 r(r+1)!

Factorials also allow telescoping. Noticing that r(r+1)! = r+1−1(r+1)! = 1r! − 1(r+1)!. Hence:

∑_r=1ⁿ r(r+1)! = 1 − 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 r(r+1)! = r+1−1(r+1)! = 1r! − 1(r+1)!。因此：

∑_r=1ⁿ r(r+1)! = 1 − 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。

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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 12 or 13 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 → ∞ 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 12 或 13 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 → ∞。

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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