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Sequences

The Dislocation Subtraction Method

The dislocation subtraction method is the method we used when deriving the summation formula of a geometric sequence. It is specially used to handle sequences formed by the product of an arithmetic sequence and a geometric sequence.

The Principle of the Method

The dislocation subtraction method

For a sequence of the form an=bn⋅cna_n = b_n \cdot c_n, where {bn}\{b_n\} is an arithmetic sequence and {cn}\{c_n\} is a geometric sequence (with common ratio qq), sum it by multiplying by the common ratio qq and then subtracting the dislocated expressions.

Basic steps:

  1. Write Sn=a1+a2+a3+⋯+anS_n = a_1 + a_2 + a_3 + \cdots + a_n
  2. Multiply both sides by the common ratio qq: qSn=a1q+a2q+a3q+⋯+anqqS_n = a_1 q + a_2 q + a_3 q + \cdots + a_n q
  3. Subtract the dislocated expressions: Sn−qSnS_n - qS_n
  4. Simplify to obtain SnS_n

Applicable Scenarios

The dislocation subtraction method applies to:

Typical examples:

  • an=n⋅2na_n = n \cdot 2^n
  • an=(2n+1)⋅3na_n = (2n+1) \cdot 3^n
  • an=n2⋅qna_n = n^2 \cdot q^n

A Classic Application: The Sum of a Geometric Sequence

Recall the derivation of the geometric sequence summation formula:

Sn=a1+a1q+a1q2+⋯+a1qn−1S_n = a_1 + a_1 q + a_1 q^2 + \cdots + a_1 q^{n-1}

Multiply both sides by qq:

qSn=a1q+a1q2+a1q3+⋯+a1qnqS_n = a_1 q + a_1 q^2 + a_1 q^3 + \cdots + a_1 q^n

Subtract the dislocated expressions:

Sn−qSn=a1−a1qnSn(1−q)=a1(1−qn)Sn=a1(1−qn)1−q\begin{aligned} S_n - qS_n &= a_1 - a_1 q^n \\ S_n(1-q) &= a_1(1-q^n) \\ S_n &= \frac{a_1(1-q^n)}{1-q} \end{aligned}

Worked Examples

Example 1: The Basic Form

Find the sum: Sn=1⋅2+2⋅22+3⋅23+⋯+n⋅2nS_n = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + n \cdot 2^n

Solution:

Sn=1⋅2+2⋅22+3⋅23+⋯+n⋅2nS_n = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + n \cdot 2^n

Multiply both sides by 2:

2Sn=1⋅22+2⋅23+3⋅24+⋯+n⋅2n+12S_n = 1 \cdot 2^2 + 2 \cdot 2^3 + 3 \cdot 2^4 + \cdots + n \cdot 2^{n+1}

Subtract the dislocated expressions:

Sn−2Sn=1⋅2+1⋅22+1⋅23+⋯+1⋅2n−n⋅2n+1−Sn=2(1+2+22+⋯+2n−1)−n⋅2n+1−Sn=2⋅2n−12−1−n⋅2n+1−Sn=2n+1−2−n⋅2n+1−Sn=(1−n)⋅2n+1−2Sn=(n−1)⋅2n+1+2\begin{aligned} S_n - 2S_n &= 1 \cdot 2 + 1 \cdot 2^2 + 1 \cdot 2^3 + \cdots + 1 \cdot 2^n - n \cdot 2^{n+1} \\ -S_n &= 2(1 + 2 + 2^2 + \cdots + 2^{n-1}) - n \cdot 2^{n+1} \\ -S_n &= 2 \cdot \frac{2^n - 1}{2 - 1} - n \cdot 2^{n+1} \\ -S_n &= 2^{n+1} - 2 - n \cdot 2^{n+1} \\ -S_n &= (1-n) \cdot 2^{n+1} - 2 \\ S_n &= (n-1) \cdot 2^{n+1} + 2 \end{aligned}

Example 2: An Arithmetic Sequence with Coefficients

Find the sum: Sn=1⋅3+3⋅32+5⋅33+⋯+(2n−1)⋅3nS_n = 1 \cdot 3 + 3 \cdot 3^2 + 5 \cdot 3^3 + \cdots + (2n-1) \cdot 3^n

Solution:

Sn=1⋅3+3⋅32+5⋅33+⋯+(2n−1)⋅3nS_n = 1 \cdot 3 + 3 \cdot 3^2 + 5 \cdot 3^3 + \cdots + (2n-1) \cdot 3^n

Multiply both sides by 3:

3Sn=1⋅32+3⋅33+5⋅34+⋯+(2n−1)⋅3n+13S_n = 1 \cdot 3^2 + 3 \cdot 3^3 + 5 \cdot 3^4 + \cdots + (2n-1) \cdot 3^{n+1}

Subtract the dislocated expressions:

Sn−3Sn=1⋅3+2⋅32+2⋅33+⋯+2⋅3n−(2n−1)⋅3n+1−2Sn=3+2(32+33+⋯+3n)−(2n−1)⋅3n+1−2Sn=3+2⋅32(3n−1−1)3−1−(2n−1)⋅3n+1−2Sn=3+32(3n−1−1)−(2n−1)⋅3n+1−2Sn=3+3n+1−9−(2n−1)⋅3n+1−2Sn=−6+(1−2n+1)⋅3n+1−2Sn=−6+(2−2n)⋅3n+1Sn=3+(n−1)⋅3n+1\begin{aligned} S_n - 3S_n &= 1 \cdot 3 + 2 \cdot 3^2 + 2 \cdot 3^3 + \cdots + 2 \cdot 3^n - (2n-1) \cdot 3^{n+1} \\ -2S_n &= 3 + 2(3^2 + 3^3 + \cdots + 3^n) - (2n-1) \cdot 3^{n+1} \\ -2S_n &= 3 + 2 \cdot \frac{3^2(3^{n-1} - 1)}{3 - 1} - (2n-1) \cdot 3^{n+1} \\ -2S_n &= 3 + 3^2(3^{n-1} - 1) - (2n-1) \cdot 3^{n+1} \\ -2S_n &= 3 + 3^{n+1} - 9 - (2n-1) \cdot 3^{n+1} \\ -2S_n &= -6 + (1 - 2n + 1) \cdot 3^{n+1} \\ -2S_n &= -6 + (2 - 2n) \cdot 3^{n+1} \\ S_n &= 3 + (n-1) \cdot 3^{n+1} \end{aligned}
Why is it called 'dislocation' subtraction?

“Dislocation” refers to the fact that the corresponding terms of the two expressions are at different positions:

The original expression: Sn=1⋅2‾+2⋅22‾+3⋅23‾+⋯+n⋅2n‾S_n = \underline{1 \cdot 2} + \underline{2 \cdot 2^2} + \underline{3 \cdot 2^3} + \cdots + \underline{n \cdot 2^n}

After multiplying by q: qSn=1⋅22‾+2⋅23‾+⋯+(n−1)⋅2n‾+n⋅2n+1‾qS_n = \quad\quad\quad \underline{1 \cdot 2^2} + \underline{2 \cdot 2^3} + \cdots + \underline{(n-1) \cdot 2^n} + \underline{n \cdot 2^{n+1}}

As you can see, every term of the second expression is “shifted back one position” relative to the first. When subtracting:

  • The first term is kept alone
  • The intermediate terms cancel in pairs (coefficients subtract, exponents stay the same)
  • The last term is kept alone

This “dislocation” is exactly where the name of the method comes from!

Practice Problems

Exercise 1

Find the sum: Sn=1⋅3+2⋅32+3⋅33+⋯+n⋅3nS_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \cdots + n \cdot 3^n

Reference Answer(2 个标签)
sequence summationdislocation subtraction

Idea: use the dislocation subtraction method.

Detailed steps:

Sn=1⋅3+2⋅32+3⋅33+⋯+n⋅3nS_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \cdots + n \cdot 3^n

3Sn=1⋅32+2⋅33+3⋅34+⋯+n⋅3n+13S_n = 1 \cdot 3^2 + 2 \cdot 3^3 + 3 \cdot 3^4 + \cdots + n \cdot 3^{n+1}

Sn−3Sn=3+32+33+⋯+3n−n⋅3n+1−2Sn=3(3n−1)3−1−n⋅3n+1−2Sn=3n+1−32−n⋅3n+1Sn=(2n−1)⋅3n+1+34\begin{aligned} S_n - 3S_n &= 3 + 3^2 + 3^3 + \cdots + 3^n - n \cdot 3^{n+1} \\ -2S_n &= \frac{3(3^n - 1)}{3 - 1} - n \cdot 3^{n+1} \\ -2S_n &= \frac{3^{n+1} - 3}{2} - n \cdot 3^{n+1} \\ S_n &= \frac{(2n-1) \cdot 3^{n+1} + 3}{4} \end{aligned}

Answer: Sn=(2n−1)⋅3n+1+34S_n = \frac{(2n-1) \cdot 3^{n+1} + 3}{4}

Exercise 2

Find the sum: Sn=1+2x+3x2+4x3+⋯+nxn−1S_n = 1 + 2x + 3x^2 + 4x^3 + \cdots + nx^{n-1} (x≠1x \neq 1)

Reference Answer(2 个标签)
sequence summationdislocation subtraction

Idea: use the dislocation subtraction method.

Detailed steps:

Sn=1+2x+3x2+4x3+⋯+nxn−1S_n = 1 + 2x + 3x^2 + 4x^3 + \cdots + nx^{n-1}

xSn=x+2x2+3x3+4x4+⋯+nxnxS_n = x + 2x^2 + 3x^3 + 4x^4 + \cdots + nx^n

Sn−xSn=1+x+x2+x3+⋯+xn−1−nxn(1−x)Sn=1−xn1−x−nxnSn=1−xn(1−x)2−nxn1−xSn=1−xn−nxn(1−x)(1−x)2Sn=1−(n+1)xn+nxn+1(1−x)2\begin{aligned} S_n - xS_n &= 1 + x + x^2 + x^3 + \cdots + x^{n-1} - nx^n \\ (1-x)S_n &= \frac{1-x^n}{1-x} - nx^n \\ S_n &= \frac{1-x^n}{(1-x)^2} - \frac{nx^n}{1-x} \\ S_n &= \frac{1 - x^n - nx^n(1-x)}{(1-x)^2} \\ S_n &= \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2} \end{aligned}

Answer: Sn=1−(n+1)xn+nxn+1(1−x)2S_n = \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2}

Exercise 3

Adapted from a postgraduate entrance examination problem

Find the sum: Sn=1⋅2+3⋅22+5⋅23+⋯+(2n−1)⋅2nS_n = 1 \cdot 2 + 3 \cdot 2^2 + 5 \cdot 2^3 + \cdots + (2n-1) \cdot 2^n

Reference Answer(2 个标签)
sequence summationdislocation subtraction

Idea: the product of the arithmetic sequence (2n−1)(2n-1) and the geometric sequence 2n2^n.

Detailed steps:

Sn=1⋅2+3⋅22+5⋅23+⋯+(2n−1)⋅2nS_n = 1 \cdot 2 + 3 \cdot 2^2 + 5 \cdot 2^3 + \cdots + (2n-1) \cdot 2^n

2Sn=1⋅22+3⋅23+5⋅24+⋯+(2n−1)⋅2n+12S_n = 1 \cdot 2^2 + 3 \cdot 2^3 + 5 \cdot 2^4 + \cdots + (2n-1) \cdot 2^{n+1}

Sn−2Sn=1⋅2+2⋅22+2⋅23+⋯+2⋅2n−(2n−1)⋅2n+1−Sn=2+2(22+23+⋯+2n)−(2n−1)⋅2n+1−Sn=2+2⋅4(2n−1−1)2−1−(2n−1)⋅2n+1−Sn=2+2n+2−8−(2n−1)⋅2n+1−Sn=−6+2n+2−(2n−1)⋅2n+1−Sn=−6+2n+1(2−2n+1)Sn=6+(2n−3)⋅2n+1\begin{aligned} S_n - 2S_n &= 1 \cdot 2 + 2 \cdot 2^2 + 2 \cdot 2^3 + \cdots + 2 \cdot 2^n - (2n-1) \cdot 2^{n+1} \\ -S_n &= 2 + 2(2^2 + 2^3 + \cdots + 2^n) - (2n-1) \cdot 2^{n+1} \\ -S_n &= 2 + 2 \cdot \frac{4(2^{n-1} - 1)}{2 - 1} - (2n-1) \cdot 2^{n+1} \\ -S_n &= 2 + 2^{n+2} - 8 - (2n-1) \cdot 2^{n+1} \\ -S_n &= -6 + 2^{n+2} - (2n-1) \cdot 2^{n+1} \\ -S_n &= -6 + 2^{n+1}(2 - 2n + 1) \\ S_n &= 6 + (2n-3) \cdot 2^{n+1} \end{aligned}

Answer: Sn=6+(2n−3)⋅2n+1S_n = 6 + (2n-3) \cdot 2^{n+1}

Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
SnS_n求和符号S sub nThe sum of the first nn terms of a sequence
qq参数quotientThe common ratio of a geometric sequence
ana_n元素符号a sub nThe nnth term of a sequence

中英对照

中文术语英文术语音标说明
错位相减法dislocation subtraction/ˌdɪsləʊˈkeɪʃən səbˈtrækʃən/The summation method of subtracting dislocated expressions after multiplying by the common ratio
等差等比乘积arithmetic-geometric product/ˌærɪθˈmetɪk-ˌdʒiːəˈmetrɪk ˈprɒdʌkt/The product of an arithmetic sequence and a geometric sequence
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