This is a beta course, so its structure, chapters, and examples may continue to change.
The Telescoping Method
The telescoping method is one of the most ingenious methods for summing sequences. Its core idea is to “split” each term into the difference of two terms, so that most terms cancel each other during the summation, leaving only the first and last few terms.
The Principle of the Method 定义
定义是对概念、术语或对象含义的精确描述。理解定义是学习任何知识领域的基础，每个概念都有其明确的定义。
Split the general term a n a_n a n a n = f ( n ) − f ( n + k ) a_n = f(n) - f(n+k) a n = f ( n ) − f ( n + k )
Basic idea :
S n = a 1 + a 2 + a 3 + ⋯ + a n = [ f ( 1 ) − f ( 2 ) ] + [ f ( 2 ) − f ( 3 ) ] + [ f ( 3 ) − f ( 4 ) ] + ⋯ + [ f ( n ) − f ( n + 1 ) ] = f ( 1 ) − f ( n + 1 ) \begin{aligned}
S_n &= a_1 + a_2 + a_3 + \cdots + a_n \\
&= [f(1) - f(2)] + [f(2) - f(3)] + [f(3) - f(4)] + \cdots + [f(n) - f(n+1)] \\
&= f(1) - f(n+1)
\end{aligned} S n = a 1 + a 2 + a 3 + ⋯ + a n = [ f ( 1 ) − f ( 2 )] + [ f ( 2 ) − f ( 3 )] + [ f ( 3 ) − f ( 4 )] + ⋯ + [ f ( n ) − f ( n + 1 )] = f ( 1 ) − f ( n + 1 ) The intermediate f ( 2 ) , f ( 3 ) , … , f ( n ) f(2), f(3), \ldots, f(n) f ( 2 ) , f ( 3 ) , … , f ( n )
Generalization :
1 n ( n + k ) = 1 k ( 1 n − 1 n + k ) \frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right) n ( n + k ) 1 = k 1 ( n 1 − n + k 1 )
1 n + n + 1 = n + 1 − n \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n} n + n + 1 1 = n + 1 − n
Derivation : multiply both the numerator and the denominator by n + 1 − n \sqrt{n+1} - \sqrt{n} n + 1 − n
1 ( 2 n − 1 ) ( 2 n + 1 ) = 1 2 ( 1 2 n − 1 − 1 2 n + 1 ) \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) ( 2 n − 1 ) ( 2 n + 1 ) 1 = 2 1 ( 2 n − 1 1 − 2 n + 1 1 )
How do we determine whether a sequence can be telescoped? Observe the characteristics of the general term:
Fractional form : the denominator is the product of two or more factorsFactor relationship : there is a regular pattern between the factors of the denominator (e.g., they differ by a constant)Try to split : see whether it can be written in the form f ( n ) − f ( n + k ) f(n) - f(n+k) f ( n ) − f ( n + k ) Technique : for 1 n ( n + k ) \frac{1}{n(n+k)} n ( n + k ) 1 1 n ( n + k ) = A n + B n + k \frac{1}{n(n+k)} = \frac{A}{n} + \frac{B}{n+k} n ( n + k ) 1 = n A + n + k B
After putting over a common denominator, compare the coefficients to find A A A B B B
Worked Examples Example 1: Basic Telescoping
Find the sum: S n = 1 1 × 2 + 1 2 × 3 + 1 3 × 4 + ⋯ + 1 n ( n + 1 ) S_n = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{n(n+1)} S n = 1 × 2 1 + 2 × 3 1 + 3 × 4 1 + ⋯ + n ( n + 1 ) 1
Solution :
Use the telescoping formula 1 n ( n + 1 ) = 1 n − 1 n + 1 \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} n ( n + 1 ) 1 = n 1 − n + 1 1
S n = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ + ( 1 n − 1 n + 1 ) = 1 1 − 1 n + 1 = n n + 1 \begin{aligned}
S_n &= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \\
&= \frac{1}{1} - \frac{1}{n+1} \\
&= \frac{n}{n+1}
\end{aligned} S n = ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + ( 3 1 − 4 1 ) + ⋯ + ( n 1 − n + 1 1 ) = 1 1 − n + 1 1 = n + 1 n Example 2: Telescoping with a Coefficient
Find the sum: S n = 1 1 × 3 + 1 3 × 5 + 1 5 × 7 + ⋯ + 1 ( 2 n − 1 ) ( 2 n + 1 ) S_n = \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \cdots + \frac{1}{(2n-1)(2n+1)} S n = 1 × 3 1 + 3 × 5 1 + 5 × 7 1 + ⋯ + ( 2 n − 1 ) ( 2 n + 1 ) 1
Solution :
Use the formula 1 ( 2 n − 1 ) ( 2 n + 1 ) = 1 2 ( 1 2 n − 1 − 1 2 n + 1 ) \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) ( 2 n − 1 ) ( 2 n + 1 ) 1 = 2 1 ( 2 n − 1 1 − 2 n + 1 1 )
S n = 1 2 [ ( 1 1 − 1 3 ) + ( 1 3 − 1 5 ) + ⋯ + ( 1 2 n − 1 − 1 2 n + 1 ) ] = 1 2 ( 1 1 − 1 2 n + 1 ) = 1 2 ⋅ 2 n 2 n + 1 = n 2 n + 1 \begin{aligned}
S_n &= \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\right] \\
&= \frac{1}{2}\left(\frac{1}{1} - \frac{1}{2n+1}\right) \\
&= \frac{1}{2} \cdot \frac{2n}{2n+1} \\
&= \frac{n}{2n+1}
\end{aligned} S n = 2 1 [ ( 1 1 − 3 1 ) + ( 3 1 − 5 1 ) + ⋯ + ( 2 n − 1 1 − 2 n + 1 1 ) ] = 2 1 ( 1 1 − 2 n + 1 1 ) = 2 1 ⋅ 2 n + 1 2 n = 2 n + 1 n Example 3: Radical Telescoping
Find the sum: S n = 1 1 + 2 + 1 2 + 3 + ⋯ + 1 n + n + 1 S_n = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{n} + \sqrt{n+1}} S n = 1 + 2 1 + 2 + 3 1 + ⋯ + n + n + 1 1
Solution :
Multiply both the numerator and the denominator by n + 1 − n \sqrt{n+1} - \sqrt{n} n + 1 − n
1 n + n + 1 = n + 1 − n ( n + 1 + n ) ( n + 1 − n ) = n + 1 − n \frac{1}{\sqrt{n} + \sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1} + \sqrt{n})(\sqrt{n+1} - \sqrt{n})} = \sqrt{n+1} - \sqrt{n} n + n + 1 1 = ( n + 1 + n ) ( n + 1 − n ) n + 1 − n = n + 1 − n
Therefore:
S n = ( 2 − 1 ) + ( 3 − 2 ) + ⋯ + ( n + 1 − n ) = n + 1 − 1 \begin{aligned}
S_n &= (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n+1} - \sqrt{n}) \\
&= \sqrt{n+1} - 1
\end{aligned} S n = ( 2 − 1 ) + ( 3 − 2 ) + ⋯ + ( n + 1 − n ) = n + 1 − 1 Practice Problems Exercise 1
Find the sum: S n = 1 2 × 4 + 1 4 × 6 + 1 6 × 8 + ⋯ + 1 2 n ( 2 n + 2 ) S_n = \frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \cdots + \frac{1}{2n(2n+2)} S n = 2 × 4 1 + 4 × 6 1 + 6 × 8 1 + ⋯ + 2 n ( 2 n + 2 ) 1
Reference Answer (2 个标签)
sequence summation telescoping
Idea : first extract the common factor, then telescope.
Detailed steps :
1 2 n ( 2 n + 2 ) = 1 2 n ⋅ 2 ( n + 1 ) = 1 4 n ( n + 1 ) = 1 4 ( 1 n − 1 n + 1 ) \frac{1}{2n(2n+2)} = \frac{1}{2n \cdot 2(n+1)} = \frac{1}{4n(n+1)} = \frac{1}{4}\left(\frac{1}{n} - \frac{1}{n+1}\right) 2 n ( 2 n + 2 ) 1 = 2 n ⋅ 2 ( n + 1 ) 1 = 4 n ( n + 1 ) 1 = 4 1 ( n 1 − n + 1 1 )
S n = 1 4 [ ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ⋯ + ( 1 n − 1 n + 1 ) ] = 1 4 ( 1 − 1 n + 1 ) = n 4 ( n + 1 ) \begin{aligned}
S_n &= \frac{1}{4}\left[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right] \\
&= \frac{1}{4}\left(1 - \frac{1}{n+1}\right) \\
&= \frac{n}{4(n+1)}
\end{aligned} S n = 4 1 [ ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + ⋯ + ( n 1 − n + 1 1 ) ] = 4 1 ( 1 − n + 1 1 ) = 4 ( n + 1 ) n Answer : S n = n 4 ( n + 1 ) S_n = \frac{n}{4(n+1)} S n = 4 ( n + 1 ) n
Exercise 2
Find the sum: S n = 1 1 × 4 + 1 4 × 7 + 1 7 × 10 + ⋯ + 1 ( 3 n − 2 ) ( 3 n + 1 ) S_n = \frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \cdots + \frac{1}{(3n-2)(3n+1)} S n = 1 × 4 1 + 4 × 7 1 + 7 × 10 1 + ⋯ + ( 3 n − 2 ) ( 3 n + 1 ) 1
Reference Answer (2 个标签)
sequence summation telescoping
Idea : the denominators differ by 3, so use the formula 1 n ( n + k ) = 1 k ( 1 n − 1 n + k ) \frac{1}{n(n+k)} = \frac{1}{k}(\frac{1}{n} - \frac{1}{n+k}) n ( n + k ) 1 = k 1 ( n 1 − n + k 1 )
Detailed steps :
1 ( 3 n − 2 ) ( 3 n + 1 ) = 1 3 ( 1 3 n − 2 − 1 3 n + 1 ) \frac{1}{(3n-2)(3n+1)} = \frac{1}{3}\left(\frac{1}{3n-2} - \frac{1}{3n+1}\right) ( 3 n − 2 ) ( 3 n + 1 ) 1 = 3 1 ( 3 n − 2 1 − 3 n + 1 1 )
S n = 1 3 [ ( 1 1 − 1 4 ) + ( 1 4 − 1 7 ) + ⋯ + ( 1 3 n − 2 − 1 3 n + 1 ) ] = 1 3 ( 1 − 1 3 n + 1 ) = n 3 n + 1 \begin{aligned}
S_n &= \frac{1}{3}\left[\left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{3n-2} - \frac{1}{3n+1}\right)\right] \\
&= \frac{1}{3}\left(1 - \frac{1}{3n+1}\right) \\
&= \frac{n}{3n+1}
\end{aligned} S n = 3 1 [ ( 1 1 − 4 1 ) + ( 4 1 − 7 1 ) + ⋯ + ( 3 n − 2 1 − 3 n + 1 1 ) ] = 3 1 ( 1 − 3 n + 1 1 ) = 3 n + 1 n Answer : S n = n 3 n + 1 S_n = \frac{n}{3n+1} S n = 3 n + 1 n
Exercise 3
Find the sum: S n = 1 1 + 3 + 1 3 + 5 + ⋯ + 1 2 n − 1 + 2 n + 1 S_n = \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \cdots + \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}} S n = 1 + 3 1 + 3 + 5 1 + ⋯ + 2 n − 1 + 2 n + 1 1
Reference Answer (2 个标签)
sequence summation telescoping
Idea : multiply both the numerator and the denominator by the conjugate.
Detailed steps :
1 2 n − 1 + 2 n + 1 = 2 n + 1 − 2 n − 1 2 \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}} = \frac{\sqrt{2n+1} - \sqrt{2n-1}}{2} 2 n − 1 + 2 n + 1 1 = 2 2 n + 1 − 2 n − 1
S n = 1 2 [ ( 3 − 1 ) + ( 5 − 3 ) + ⋯ + ( 2 n + 1 − 2 n − 1 ) ] = 1 2 ( 2 n + 1 − 1 ) \begin{aligned}
S_n &= \frac{1}{2}[(\sqrt{3} - \sqrt{1}) + (\sqrt{5} - \sqrt{3}) + \cdots + (\sqrt{2n+1} - \sqrt{2n-1})] \\
&= \frac{1}{2}(\sqrt{2n+1} - 1)
\end{aligned} S n = 2 1 [( 3 − 1 ) + ( 5 − 3 ) + ⋯ + ( 2 n + 1 − 2 n − 1 )] = 2 1 ( 2 n + 1 − 1 ) Answer : S n = 2 n + 1 − 1 2 S_n = \frac{\sqrt{2n+1} - 1}{2} S n = 2 2 n + 1 − 1
Exercise 4
Adapted from a postgraduate entrance examination problem
Find the sum: S n = 1 1 × 2 × 3 + 1 2 × 3 × 4 + ⋯ + 1 n ( n + 1 ) ( n + 2 ) S_n = \frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \cdots + \frac{1}{n(n+1)(n+2)} S n = 1 × 2 × 3 1 + 2 × 3 × 4 1 + ⋯ + n ( n + 1 ) ( n + 2 ) 1
Reference Answer (2 个标签)
sequence summation telescoping
Idea : for a product of three consecutive integers, telescoping needs to be done twice.
Detailed steps :
1 n ( n + 1 ) ( n + 2 ) = 1 2 [ 1 n ( n + 1 ) − 1 ( n + 1 ) ( n + 2 ) ] \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right] n ( n + 1 ) ( n + 2 ) 1 = 2 1 [ n ( n + 1 ) 1 − ( n + 1 ) ( n + 2 ) 1 ]
S n = 1 2 [ ( 1 1 × 2 − 1 2 × 3 ) + ( 1 2 × 3 − 1 3 × 4 ) + ⋯ ] = 1 2 ( 1 1 × 2 − 1 ( n + 1 ) ( n + 2 ) ) = 1 2 ( 1 2 − 1 ( n + 1 ) ( n + 2 ) ) = 1 4 − 1 2 ( n + 1 ) ( n + 2 ) = n ( n + 3 ) 4 ( n + 1 ) ( n + 2 ) \begin{aligned}
S_n &= \frac{1}{2}\left[\left(\frac{1}{1 \times 2} - \frac{1}{2 \times 3}\right) + \left(\frac{1}{2 \times 3} - \frac{1}{3 \times 4}\right) + \cdots\right] \\
&= \frac{1}{2}\left(\frac{1}{1 \times 2} - \frac{1}{(n+1)(n+2)}\right) \\
&= \frac{1}{2}\left(\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right) \\
&= \frac{1}{4} - \frac{1}{2(n+1)(n+2)} \\
&= \frac{n(n+3)}{4(n+1)(n+2)}
\end{aligned} S n = 2 1 [ ( 1 × 2 1 − 2 × 3 1 ) + ( 2 × 3 1 − 3 × 4 1 ) + ⋯ ] = 2 1 ( 1 × 2 1 − ( n + 1 ) ( n + 2 ) 1 ) = 2 1 ( 2 1 − ( n + 1 ) ( n + 2 ) 1 ) = 4 1 − 2 ( n + 1 ) ( n + 2 ) 1 = 4 ( n + 1 ) ( n + 2 ) n ( n + 3 ) Answer : S n = n ( n + 3 ) 4 ( n + 1 ) ( n + 2 ) S_n = \frac{n(n+3)}{4(n+1)(n+2)} S n = 4 ( n + 1 ) ( n + 2 ) n ( n + 3 )
Summary Symbols Used in This Article
符号 类型 读音/说明 在本文中的含义 S n S_n S n 求和符号 S sub n The sum of the first n n n f ( n ) f(n) f ( n ) 函数符号 f of n The function form after telescoping k k k 常数 k The telescoping interval
中英对照
中文术语 英文术语 音标 说明 裂项相消法 telescoping series /ˈtelɪskəʊpɪŋ ˈsɪəriːz/ The summation method in which intermediate terms cancel by splitting terms 通分 common denominator /ˈkɒmən dɪˈnɒmɪneɪtə/ Reducing fractions to a common denominator 共轭 conjugate /ˈkɒndʒʊɡət/ Such as a + b \sqrt{a} + \sqrt{b} a + b a − b \sqrt{a} - \sqrt{b} a − b
