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

📚 Summation of Series by the Method of Differences | 差分法求和

The method of differences is a powerful technique for summing finite series whose terms can be expressed as the difference of two consecutive terms of another sequence. Often called "telescoping", this method simplifies a long summation into just a few evaluations at the boundaries.

差分法是求有限级数和的一种强大技巧，适用于项可以表示为另一个数列相邻两项之差的级数。它通常被称为“裂项相消”，能将长求和简化为仅在边界处的若干计算。

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1. The Core Idea | 核心思想

Suppose we can write the general term u_k of a series as u_k = v_k+1 - v_k for some sequence v_k. Then the sum from k=1 to n becomes:

假设我们能将级数的通项 u_k 写成 u_k = v_k+1 - v_k，其中 v_k 是某个数列。则从 k=1 到 n 的和变为：

∑_k=1ⁿ (v_k+1 - v_k) = v_n+1 - v₁

All intermediate terms cancel in pairs. This is why the method is also called "telescoping" – the sum collapses like a telescope.

所有中间项成对相消。这也是为什么该方法也叫“望远镜求和”——求和如同望远镜一样收缩。

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2. When to Use It | 何时使用

The method is particularly effective when the general term is a rational function whose denominator factors into consecutive integers, or when it can be expressed as a partial fraction difference. Typical forms include:

当通项是分母可分解为连续整数的有理函数，或者可以表示为部分分式之差时，该方法尤为有效。常见形式包括：

• 1k(k+1) – decomposes to 1k - 1k+1
• 1k(k+1)(k+2) – can be split using partial fractions
• 1(2k-1)(2k+1) – useful for series of reciprocals of odd products

• 1k(k+1) – 分解为 1k - 1k+1
• 1k(k+1)(k+2) – 可用部分分式拆分
• 1(2k-1)(2k+1) – 适用于奇数乘积倒数级数

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3. Partial Fractions and Differences | 部分分式与差分

Often the given term is not already in the form v_k+1 - v_k. We first carry out a partial fraction decomposition. For example:

通常给出的项并不直接是 v_k+1 - v_k 的形式，我们需要先进行部分分式分解。例如：

1k(k+2) = 12( 1k - 1k+2 )

This introduces a “gap” of 2 between the terms, so when summing, terms do not all cancel; only every second term cancels, leaving two boundary sums.

这里出现了差距为2的项，因此在求和时并非所有项都相消，而是每隔一项相消，剩余两个边界和。

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4. Worked Example 1: Basic Telescoping | 例1：基础裂项

Find the sum S_n = ∑_k=1ⁿ 1k(k+1).

求 S_n = ∑_k=1ⁿ 1k(k+1) 的和。

We decompose the term:

我们对通项进行分解：

1k(k+1) = 1k - 1k+1

Then the sum is:

于是和为：

S_n = (1 - 12) + (12 - 13) + ⋯ + (1n - 1n+1) = 1 - 1n+1

Hence S_n = nn+1.

因此 S_n = nn+1。

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5. Worked Example 2: Non-unit Numerator | 例2：非单位分子

Sum the series ∑_k=1ⁿ 2k(k+2).

求级数 ∑_k=1ⁿ 2k(k+2) 的和。

Decompose:

分解：

2k(k+2) = 1k - 1k+2

Now write out the terms:

现在写出各项：

S_n = (1 - 13) + (12 - 14) + (13 - 15) + ⋯

After cancellation, the remaining terms are 1 + 12 - 1n+1 - 1n+2. Therefore:

相消后，剩余项为 1 + 12 - 1n+1 - 1n+2。因此：

S_n = 32 - 1n+1 - 1n+2

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6. Systematic Cancellation Table | 系统化的相消表

For more complex decompositions, it helps to list the first few and the last few terms explicitly. Consider a gap of d: the remaining terms are the first d terms of v_k and the last d terms of v_k.

对于更复杂的分解，明确列出前几项和后几项会有帮助。设差距为 d，则剩余项为 v_k 的前 d 项和后 d 项。

Gap d	Remaining positive terms	Remaining negative terms
1	v1	vn+1
2	v1 + v2	vn+1 + vn+2
3	v1 + v2 + v3	vn+1 + vn+2 + vn+3

Always verify the pattern by writing three or four terms explicitly before jumping to the final formula.

在直接写出最终公式前，务必先写出三四项来验证规律。

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7. When the Term Is Not Already a Difference | 当通项本身不是差的形式

Sometimes the term u_k can be written as v_k - v_k+1 (reverse order) or as v_k - v_k+m with a constant m. In such cases, the same telescoping logic applies, but careful attention is needed to the sign and the boundaries.

有时通项 u_k 可以写成 v_k - v_k+1（反向）或 v_k - v_k+m（其中 m 为常数）。此时同样的裂项逻辑适用，但需要仔细注意符号和边界。

For example:

例如：

1(2k-1)(2k+1) = 12( 12k-1 - 12k+1 )

Here v_k = 12k-1, and the gap is 2. The sum is:

这里 v_k = 12k-1，差距为2。其和为：

S_n = 12( 1 - 12n+1 )

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8. Combining with Standard Formulas | 与标准公式结合

The method of differences can also be used to derive summation formulas for polynomial sequences. For instance, since (k+1)³ - k³ = 3k² + 3k + 1, summing from 1 to n gives a direct proof of the sum of squares formula.

差分法也可用于推导多项式数列的求和公式。例如，因为 (k+1)³ - k³ = 3k² + 3k + 1，从1到n求和即可直接证明平方和公式。

Summing both sides:

两边求和：

(n+1)³ - 1 = 3∑_k=1ⁿ k² + 3∑_k=1ⁿ k + n

Rearranging gives:

整理得到：

∑_k=1ⁿ k² = n(n+1)(2n+1)6

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9. Common Pitfalls | 常见陷阱

Students often make errors in the constant factor obtained from partial fractions. For example, writing 1k(k+2) = 1k - 1k+2 is incorrect; the correct decomposition includes a factor of 12.

学生常在部分分式中的常数因子出错。例如，把 1k(k+2) = 1k - 1k+2 写成这样是错误的；正确的分解包含因子 12。

Another common mistake is failing to identify which terms actually cancel. Always write out at least five terms: the first two and the last two, to see the pattern clearly.

另一个常见错误是未能识别哪些项真正相消。务必至少写出五项：前两项和后两项，以看清规律。

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10. Practice Questions | 练习

Try the following on your own:

请自行尝试以下题目：

1. ∑_k=1ⁿ 1(3k-1)(3k+2) – hint: decompose into a difference with gap 3.
2. ∑_k=1ⁿ k(k+1)! – hint: express k = (k+1)-1 and use factorial differences.
3. ∑_k=1ⁿ ln(1+1k) – use logarithm laws to telescope.

1. ∑_k=1ⁿ 1(3k-1)(3k+2) – 提示：分解为差距为3的差。
2. ∑_k=1ⁿ k(k+1)! – 提示：将 k = (k+1)-1 并利用阶乘差分。
3. ∑_k=1ⁿ ln(1+1k) – 利用对数运算法则裂项。

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11. Summary | 总结

The method of differences is a cornerstone of A-level series summation. Master it by practicing partial fraction decomposition and by meticulously tracking which terms survive.

差分法是A-Level级数求和的基石。通过练习部分分式分解并仔细追踪剩余项来掌握它。

Remember the three essential steps:

记住三个关键步骤：

• Decompose the general term into a difference of two terms.
• Write out the sum explicitly for the first few and last few terms.
• Cancel matching pairs and simplify the boundary terms.

• 将通项分解为两项之差。
• 明确写出前几项和后几项的求和。
• 消去成对项并化简边界项。

With practice, telescoping sums become intuitive and quick.

多加练习，裂项求和将变得直观而快速。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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