This is a beta course, so its structure, chapters, and examples may continue to change.
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 For a sequence of the form a n = b n ⋅ c n a_n = b_n \cdot c_n a n = b n ⋅ c n { b n } \{b_n\} { b n } { c n } \{c_n\} { c n } q q q q q q
Basic steps :
Write S n = a 1 + a 2 + a 3 + ⋯ + a n S_n = a_1 + a_2 + a_3 + \cdots + a_n S n = a 1 + a 2 + a 3 + ⋯ + a n
Multiply both sides by the common ratio q q q q S n = a 1 q + a 2 q + a 3 q + ⋯ + a n q qS_n = a_1 q + a_2 q + a_3 q + \cdots + a_n q q S n = a 1 q + a 2 q + a 3 q + ⋯ + a n q
Subtract the dislocated expressions: S n − q S n S_n - qS_n S n − q S n
Simplify to obtain S n S_n S n
Applicable Scenarios The dislocation subtraction method applies to:
General term form : a n = ( a n + b ) ⋅ q n a_n = (an + b) \cdot q^n a n = ( an + b ) ⋅ q n a n = f ( n ) ⋅ q n a_n = f(n) \cdot q^n a n = f ( n ) ⋅ q n
where f ( n ) f(n) f ( n ) n n n q q q q ≠ 1 q \neq 1 q = 1
Typical examples :
a n = n ⋅ 2 n a_n = n \cdot 2^n a n = n ⋅ 2 n a n = ( 2 n + 1 ) ⋅ 3 n a_n = (2n+1) \cdot 3^n a n = ( 2 n + 1 ) ⋅ 3 n a n = n 2 ⋅ q n a_n = n^2 \cdot q^n a n = n 2 ⋅ q n
A Classic Application: The Sum of a Geometric Sequence Recall the derivation of the geometric sequence summation formula:
S n = a 1 + a 1 q + a 1 q 2 + ⋯ + a 1 q n − 1 S_n = a_1 + a_1 q + a_1 q^2 + \cdots + a_1 q^{n-1} S n = a 1 + a 1 q + a 1 q 2 + ⋯ + a 1 q n − 1
Multiply both sides by q q q
q S n = a 1 q + a 1 q 2 + a 1 q 3 + ⋯ + a 1 q n qS_n = a_1 q + a_1 q^2 + a_1 q^3 + \cdots + a_1 q^n q S n = a 1 q + a 1 q 2 + a 1 q 3 + ⋯ + a 1 q n
Subtract the dislocated expressions:
S n − q S n = a 1 − a 1 q n S n ( 1 − q ) = a 1 ( 1 − q n ) S n = a 1 ( 1 − q n ) 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} S n − q S n S n ( 1 − q ) S n = a 1 − a 1 q n = a 1 ( 1 − q n ) = 1 − q a 1 ( 1 − q n ) Worked Examples Find the sum: S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 3 ⋅ 2 3 + ⋯ + n ⋅ 2 n S_n = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + n \cdot 2^n S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 3 ⋅ 2 3 + ⋯ + n ⋅ 2 n
Solution :
S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 3 ⋅ 2 3 + ⋯ + n ⋅ 2 n S_n = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + n \cdot 2^n S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 3 ⋅ 2 3 + ⋯ + n ⋅ 2 n
Multiply both sides by 2:
2 S n = 1 ⋅ 2 2 + 2 ⋅ 2 3 + 3 ⋅ 2 4 + ⋯ + n ⋅ 2 n + 1 2S_n = 1 \cdot 2^2 + 2 \cdot 2^3 + 3 \cdot 2^4 + \cdots + n \cdot 2^{n+1} 2 S n = 1 ⋅ 2 2 + 2 ⋅ 2 3 + 3 ⋅ 2 4 + ⋯ + n ⋅ 2 n + 1
Subtract the dislocated expressions:
S n − 2 S n = 1 ⋅ 2 + 1 ⋅ 2 2 + 1 ⋅ 2 3 + ⋯ + 1 ⋅ 2 n − n ⋅ 2 n + 1 − S n = 2 ( 1 + 2 + 2 2 + ⋯ + 2 n − 1 ) − n ⋅ 2 n + 1 − S n = 2 ⋅ 2 n − 1 2 − 1 − n ⋅ 2 n + 1 − S n = 2 n + 1 − 2 − n ⋅ 2 n + 1 − S n = ( 1 − n ) ⋅ 2 n + 1 − 2 S n = ( n − 1 ) ⋅ 2 n + 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} S n − 2 S n − S n − S n − S n − S n S n = 1 ⋅ 2 + 1 ⋅ 2 2 + 1 ⋅ 2 3 + ⋯ + 1 ⋅ 2 n − n ⋅ 2 n + 1 = 2 ( 1 + 2 + 2 2 + ⋯ + 2 n − 1 ) − n ⋅ 2 n + 1 = 2 ⋅ 2 − 1 2 n − 1 − n ⋅ 2 n + 1 = 2 n + 1 − 2 − n ⋅ 2 n + 1 = ( 1 − n ) ⋅ 2 n + 1 − 2 = ( n − 1 ) ⋅ 2 n + 1 + 2 Example 2: An Arithmetic Sequence with Coefficients
Find the sum: S n = 1 ⋅ 3 + 3 ⋅ 3 2 + 5 ⋅ 3 3 + ⋯ + ( 2 n − 1 ) ⋅ 3 n S_n = 1 \cdot 3 + 3 \cdot 3^2 + 5 \cdot 3^3 + \cdots + (2n-1) \cdot 3^n S n = 1 ⋅ 3 + 3 ⋅ 3 2 + 5 ⋅ 3 3 + ⋯ + ( 2 n − 1 ) ⋅ 3 n
Solution :
S n = 1 ⋅ 3 + 3 ⋅ 3 2 + 5 ⋅ 3 3 + ⋯ + ( 2 n − 1 ) ⋅ 3 n S_n = 1 \cdot 3 + 3 \cdot 3^2 + 5 \cdot 3^3 + \cdots + (2n-1) \cdot 3^n S n = 1 ⋅ 3 + 3 ⋅ 3 2 + 5 ⋅ 3 3 + ⋯ + ( 2 n − 1 ) ⋅ 3 n
Multiply both sides by 3:
3 S n = 1 ⋅ 3 2 + 3 ⋅ 3 3 + 5 ⋅ 3 4 + ⋯ + ( 2 n − 1 ) ⋅ 3 n + 1 3S_n = 1 \cdot 3^2 + 3 \cdot 3^3 + 5 \cdot 3^4 + \cdots + (2n-1) \cdot 3^{n+1} 3 S n = 1 ⋅ 3 2 + 3 ⋅ 3 3 + 5 ⋅ 3 4 + ⋯ + ( 2 n − 1 ) ⋅ 3 n + 1
Subtract the dislocated expressions:
S n − 3 S n = 1 ⋅ 3 + 2 ⋅ 3 2 + 2 ⋅ 3 3 + ⋯ + 2 ⋅ 3 n − ( 2 n − 1 ) ⋅ 3 n + 1 − 2 S n = 3 + 2 ( 3 2 + 3 3 + ⋯ + 3 n ) − ( 2 n − 1 ) ⋅ 3 n + 1 − 2 S n = 3 + 2 ⋅ 3 2 ( 3 n − 1 − 1 ) 3 − 1 − ( 2 n − 1 ) ⋅ 3 n + 1 − 2 S n = 3 + 3 2 ( 3 n − 1 − 1 ) − ( 2 n − 1 ) ⋅ 3 n + 1 − 2 S n = 3 + 3 n + 1 − 9 − ( 2 n − 1 ) ⋅ 3 n + 1 − 2 S n = − 6 + ( 1 − 2 n + 1 ) ⋅ 3 n + 1 − 2 S n = − 6 + ( 2 − 2 n ) ⋅ 3 n + 1 S n = 3 + ( n − 1 ) ⋅ 3 n + 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} S n − 3 S n − 2 S n − 2 S n − 2 S n − 2 S n − 2 S n − 2 S n S n = 1 ⋅ 3 + 2 ⋅ 3 2 + 2 ⋅ 3 3 + ⋯ + 2 ⋅ 3 n − ( 2 n − 1 ) ⋅ 3 n + 1 = 3 + 2 ( 3 2 + 3 3 + ⋯ + 3 n ) − ( 2 n − 1 ) ⋅ 3 n + 1 = 3 + 2 ⋅ 3 − 1 3 2 ( 3 n − 1 − 1 ) − ( 2 n − 1 ) ⋅ 3 n + 1 = 3 + 3 2 ( 3 n − 1 − 1 ) − ( 2 n − 1 ) ⋅ 3 n + 1 = 3 + 3 n + 1 − 9 − ( 2 n − 1 ) ⋅ 3 n + 1 = − 6 + ( 1 − 2 n + 1 ) ⋅ 3 n + 1 = − 6 + ( 2 − 2 n ) ⋅ 3 n + 1 = 3 + ( n − 1 ) ⋅ 3 n + 1 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 :
S n = 1 ⋅ 2 ‾ + 2 ⋅ 2 2 ‾ + 3 ⋅ 2 3 ‾ + ⋯ + n ⋅ 2 n ‾ S_n = \underline{1 \cdot 2} + \underline{2 \cdot 2^2} + \underline{3 \cdot 2^3} + \cdots + \underline{n \cdot 2^n} S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 3 ⋅ 2 3 + ⋯ + n ⋅ 2 n
After multiplying by q :
q S n = 1 ⋅ 2 2 ‾ + 2 ⋅ 2 3 ‾ + ⋯ + ( n − 1 ) ⋅ 2 n ‾ + n ⋅ 2 n + 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}} q S n = 1 ⋅ 2 2 + 2 ⋅ 2 3 + ⋯ + ( n − 1 ) ⋅ 2 n + n ⋅ 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: S n = 1 ⋅ 3 + 2 ⋅ 3 2 + 3 ⋅ 3 3 + ⋯ + n ⋅ 3 n S_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \cdots + n \cdot 3^n S n = 1 ⋅ 3 + 2 ⋅ 3 2 + 3 ⋅ 3 3 + ⋯ + n ⋅ 3 n
Reference Answer sequence summation dislocation subtraction
Approach : use the dislocation subtraction method.
Steps :
S n = 1 ⋅ 3 + 2 ⋅ 3 2 + 3 ⋅ 3 3 + ⋯ + n ⋅ 3 n S_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \cdots + n \cdot 3^n S n = 1 ⋅ 3 + 2 ⋅ 3 2 + 3 ⋅ 3 3 + ⋯ + n ⋅ 3 n
3 S n = 1 ⋅ 3 2 + 2 ⋅ 3 3 + 3 ⋅ 3 4 + ⋯ + n ⋅ 3 n + 1 3S_n = 1 \cdot 3^2 + 2 \cdot 3^3 + 3 \cdot 3^4 + \cdots + n \cdot 3^{n+1} 3 S n = 1 ⋅ 3 2 + 2 ⋅ 3 3 + 3 ⋅ 3 4 + ⋯ + n ⋅ 3 n + 1
S n − 3 S n = 3 + 3 2 + 3 3 + ⋯ + 3 n − n ⋅ 3 n + 1 − 2 S n = 3 ( 3 n − 1 ) 3 − 1 − n ⋅ 3 n + 1 − 2 S n = 3 n + 1 − 3 2 − n ⋅ 3 n + 1 S n = ( 2 n − 1 ) ⋅ 3 n + 1 + 3 4 \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} S n − 3 S n − 2 S n − 2 S n S n = 3 + 3 2 + 3 3 + ⋯ + 3 n − n ⋅ 3 n + 1 = 3 − 1 3 ( 3 n − 1 ) − n ⋅ 3 n + 1 = 2 3 n + 1 − 3 − n ⋅ 3 n + 1 = 4 ( 2 n − 1 ) ⋅ 3 n + 1 + 3 Answer : S n = ( 2 n − 1 ) ⋅ 3 n + 1 + 3 4 S_n = \frac{(2n-1) \cdot 3^{n+1} + 3}{4} S n = 4 ( 2 n − 1 ) ⋅ 3 n + 1 + 3
Exercise 2
Find the sum: S n = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ + n x n − 1 S_n = 1 + 2x + 3x^2 + 4x^3 + \cdots + nx^{n-1} S n = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ + n x n − 1 x ≠ 1 x \neq 1 x = 1
Reference Answer sequence summation dislocation subtraction
Approach : use the dislocation subtraction method.
Steps :
S n = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ + n x n − 1 S_n = 1 + 2x + 3x^2 + 4x^3 + \cdots + nx^{n-1} S n = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ + n x n − 1
x S n = x + 2 x 2 + 3 x 3 + 4 x 4 + ⋯ + n x n xS_n = x + 2x^2 + 3x^3 + 4x^4 + \cdots + nx^n x S n = x + 2 x 2 + 3 x 3 + 4 x 4 + ⋯ + n x n
S n − x S n = 1 + x + x 2 + x 3 + ⋯ + x n − 1 − n x n ( 1 − x ) S n = 1 − x n 1 − x − n x n S n = 1 − x n ( 1 − x ) 2 − n x n 1 − x S n = 1 − x n − n x n ( 1 − x ) ( 1 − x ) 2 S n = 1 − ( n + 1 ) x n + n x n + 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} S n − x S n ( 1 − x ) S n S n S n S n = 1 + x + x 2 + x 3 + ⋯ + x n − 1 − n x n = 1 − x 1 − x n − n x n = ( 1 − x ) 2 1 − x n − 1 − x n x n = ( 1 − x ) 2 1 − x n − n x n ( 1 − x ) = ( 1 − x ) 2 1 − ( n + 1 ) x n + n x n + 1 Answer : S n = 1 − ( n + 1 ) x n + n x n + 1 ( 1 − x ) 2 S_n = \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2} S n = ( 1 − x ) 2 1 − ( n + 1 ) x n + n x n + 1
Exercise 3
Find the sum: S n = 1 ⋅ 2 + 3 ⋅ 2 2 + 5 ⋅ 2 3 + ⋯ + ( 2 n − 1 ) ⋅ 2 n S_n = 1 \cdot 2 + 3 \cdot 2^2 + 5 \cdot 2^3 + \cdots + (2n-1) \cdot 2^n S n = 1 ⋅ 2 + 3 ⋅ 2 2 + 5 ⋅ 2 3 + ⋯ + ( 2 n − 1 ) ⋅ 2 n
Reference Answer sequence summation dislocation subtraction
Approach : the product of the arithmetic sequence ( 2 n − 1 ) (2n-1) ( 2 n − 1 ) 2 n 2^n 2 n
Steps :
S n = 1 ⋅ 2 + 3 ⋅ 2 2 + 5 ⋅ 2 3 + ⋯ + ( 2 n − 1 ) ⋅ 2 n S_n = 1 \cdot 2 + 3 \cdot 2^2 + 5 \cdot 2^3 + \cdots + (2n-1) \cdot 2^n S n = 1 ⋅ 2 + 3 ⋅ 2 2 + 5 ⋅ 2 3 + ⋯ + ( 2 n − 1 ) ⋅ 2 n
2 S n = 1 ⋅ 2 2 + 3 ⋅ 2 3 + 5 ⋅ 2 4 + ⋯ + ( 2 n − 1 ) ⋅ 2 n + 1 2S_n = 1 \cdot 2^2 + 3 \cdot 2^3 + 5 \cdot 2^4 + \cdots + (2n-1) \cdot 2^{n+1} 2 S n = 1 ⋅ 2 2 + 3 ⋅ 2 3 + 5 ⋅ 2 4 + ⋯ + ( 2 n − 1 ) ⋅ 2 n + 1
S n − 2 S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 2 ⋅ 2 3 + ⋯ + 2 ⋅ 2 n − ( 2 n − 1 ) ⋅ 2 n + 1 − S n = 2 + 2 ( 2 2 + 2 3 + ⋯ + 2 n ) − ( 2 n − 1 ) ⋅ 2 n + 1 − S n = 2 + 2 ⋅ 4 ( 2 n − 1 − 1 ) 2 − 1 − ( 2 n − 1 ) ⋅ 2 n + 1 − S n = 2 + 2 n + 2 − 8 − ( 2 n − 1 ) ⋅ 2 n + 1 − S n = − 6 + 2 n + 2 − ( 2 n − 1 ) ⋅ 2 n + 1 − S n = − 6 + 2 n + 1 ( 2 − 2 n + 1 ) S n = 6 + ( 2 n − 3 ) ⋅ 2 n + 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} S n − 2 S n − S n − S n − S n − S n − S n S n = 1 ⋅ 2 + 2 ⋅ 2 2 + 2 ⋅ 2 3 + ⋯ + 2 ⋅ 2 n − ( 2 n − 1 ) ⋅ 2 n + 1 = 2 + 2 ( 2 2 + 2 3 + ⋯ + 2 n ) − ( 2 n − 1 ) ⋅ 2 n + 1 = 2 + 2 ⋅ 2 − 1 4 ( 2 n − 1 − 1 ) − ( 2 n − 1 ) ⋅ 2 n + 1 = 2 + 2 n + 2 − 8 − ( 2 n − 1 ) ⋅ 2 n + 1 = − 6 + 2 n + 2 − ( 2 n − 1 ) ⋅ 2 n + 1 = − 6 + 2 n + 1 ( 2 − 2 n + 1 ) = 6 + ( 2 n − 3 ) ⋅ 2 n + 1 Answer : S n = 6 + ( 2 n − 3 ) ⋅ 2 n + 1 S_n = 6 + (2n-3) \cdot 2^{n+1} S n = 6 + ( 2 n − 3 ) ⋅ 2 n + 1
Summary Symbols Used in This Article
Symbol Type Pronunciation/Explanation Meaning in This Article S n S_n S n Summation symbol S sub n The sum of the first n n n q q q Parameter quotient The common ratio of a geometric sequence a n a_n a n Element symbol a sub n The n n n
