Kuzee
Sequences

The Grouping Summation Method

The grouping summation method is the most direct method for handling composite sequences. When the general term of a sequence can be decomposed into the sum of several simple sequences, we can sum each group separately and then add the results.

The Principle of the Method

The grouping summation method

Decompose the general term ana_n of a sequence into the sum of several simple sequences an=bn+cn+⋯a_n = b_n + c_n + \cdots, sum each group separately, and then add them: Sn=∑an=∑bn+∑cn+⋯S_n = \sum a_n = \sum b_n + \sum c_n + \cdots

Core idea: break the whole into parts and conquer each one.

Applicable Scenarios

The grouping summation method applies to the following cases:

  1. Arithmetic + geometric: an=(an+b)+cqna_n = (an + b) + cq^n
  2. Polynomial + exponential: an=n2+2na_n = n^2 + 2^n
  3. Piecewise definitions: odd-indexed and even-indexed terms follow different rules

Worked Examples

Example 1: Arithmetic + Geometric

Find the sum: Sn=(1+2)+(2+4)+(3+8)+⋯+(n+2n)S_n = (1 + 2) + (2 + 4) + (3 + 8) + \cdots + (n + 2^n)

Solution:

Split the sequence into two groups:

Sn=(1+2+3+⋯+n)+(2+4+8+⋯+2n)S_n = (1 + 2 + 3 + \cdots + n) + (2 + 4 + 8 + \cdots + 2^n)

The first group is an arithmetic sequence and the second is a geometric sequence:

Sn=n(n+1)2+2(2n−1)2−1=n(n+1)2+2n+1−2\begin{aligned} S_n &= \frac{n(n+1)}{2} + \frac{2(2^n - 1)}{2 - 1} \\ &= \frac{n(n+1)}{2} + 2^{n+1} - 2 \end{aligned}

Example 2: Polynomial + Exponential

Find the sum: Sn=(1+31)+(4+32)+(9+33)+⋯+(n2+3n)S_n = (1 + 3^1) + (4 + 3^2) + (9 + 3^3) + \cdots + (n^2 + 3^n)

Solution:

Grouping:

Sn=(1+4+9+⋯+n2)+(3+32+33+⋯+3n)S_n = (1 + 4 + 9 + \cdots + n^2) + (3 + 3^2 + 3^3 + \cdots + 3^n)

The first group is a sum of squares and the second is a geometric sequence:

Sn=n(n+1)(2n+1)6+3(3n−1)3−1=n(n+1)(2n+1)6+3n+1−32\begin{aligned} S_n &= \frac{n(n+1)(2n+1)}{6} + \frac{3(3^n - 1)}{3 - 1} \\ &= \frac{n(n+1)(2n+1)}{6} + \frac{3^{n+1} - 3}{2} \end{aligned}

Example 3: Grouping by Odd and Even Terms

The sequence {an}\{a_n\} satisfies: the odd-indexed terms form the arithmetic sequence 1,3,5,…1, 3, 5, \ldots, and the even-indexed terms form the geometric sequence 2,4,8,…2, 4, 8, \ldots. Find the sum of the first 2n2n terms.

Solution:

S2n=(a1+a3+⋯+a2n−1)+(a2+a4+⋯+a2n)=[1+3+5+⋯+(2n−1)]+[2+4+8+⋯+2n]=n2+2(2n−1)2−1=n2+2n+1−2\begin{aligned} S_{2n} &= (a_1 + a_3 + \cdots + a_{2n-1}) + (a_2 + a_4 + \cdots + a_{2n}) \\ &= [1 + 3 + 5 + \cdots + (2n-1)] + [2 + 4 + 8 + \cdots + 2^n] \\ &= n^2 + \frac{2(2^n - 1)}{2 - 1} \\ &= n^2 + 2^{n+1} - 2 \end{aligned}
What is the key to the grouping summation method?

The key lies in correctly identifying and grouping:

  1. Observe the structure of the general term: see whether it can be split into several independent parts
  2. Identify the type of each group: determine whether each group is arithmetic, geometric, or another known type
  3. Sum each group separately: use the corresponding summation formula for each group
  4. Combine the results: add the sums of all the groups

Common mistakes:

  • Incomplete grouping, leaving complex terms behind
  • Forgetting a group after grouping
  • Confusing the summation formulas of different groups

Practice Problems

Exercise 1

Find the sum: Sn=(1+1)+(2+3)+(3+9)+⋯+(n+3n)S_n = (1 + 1) + (2 + 3) + (3 + 9) + \cdots + (n + 3^n)

Reference Answer(2 个标签)
sequence summationgrouping method

Idea: split into an arithmetic sequence group and a geometric sequence group.

Detailed steps:

Sn=(1+2+3+⋯+n)+(1+3+9+⋯+3n)S_n = (1 + 2 + 3 + \cdots + n) + (1 + 3 + 9 + \cdots + 3^n)

Sn=n(n+1)2+1×(3n−1)3−1=n(n+1)2+3n−12=n(n+1)+3n−12\begin{aligned} S_n &= \frac{n(n+1)}{2} + \frac{1 \times (3^n - 1)}{3 - 1} \\ &= \frac{n(n+1)}{2} + \frac{3^n - 1}{2} \\ &= \frac{n(n+1) + 3^n - 1}{2} \end{aligned}

Answer: Sn=n(n+1)+3n−12S_n = \frac{n(n+1) + 3^n - 1}{2}

Exercise 2

Find the sum: Sn=2+22+2+23+2+24+⋯S_n = 2 + 2^2 + 2 + 2^3 + 2 + 2^4 + \cdots (2n2n terms in total)

Reference Answer(2 个标签)
sequence summationgrouping method

Idea: the odd-indexed terms are all 2, and the even-indexed terms form a geometric sequence.

Detailed steps:

Odd-indexed terms: 2,2,2,…2, 2, 2, \ldots (nn copies of 2)

Even-indexed terms: 22,23,24,…,2n+12^2, 2^3, 2^4, \ldots, 2^{n+1} (a geometric sequence)

S2n=2n+22(2n−1)2−1=2n+4(2n−1)=2n+2n+2−4\begin{aligned} S_{2n} &= 2n + \frac{2^2(2^n - 1)}{2 - 1} \\ &= 2n + 4(2^n - 1) \\ &= 2n + 2^{n+2} - 4 \end{aligned}

Answer: S2n=2n+2n+2−4S_{2n} = 2n + 2^{n+2} - 4

Exercise 3

Find the sum: Sn=(2×1+1)+(2×2+2)+(2×3+4)+⋯+(2n+2n)S_n = (2 \times 1 + 1) + (2 \times 2 + 2) + (2 \times 3 + 4) + \cdots + (2n + 2^n)

Reference Answer(2 个标签)
sequence summationgrouping method

Idea: expand and group.

Detailed steps:

Sn=(2+4+6+⋯+2n)+(1+2+4+⋯+2n)S_n = (2 + 4 + 6 + \cdots + 2n) + (1 + 2 + 4 + \cdots + 2^n)

The first group is an arithmetic sequence with first term 2 and common difference 2:

∑k=1n2k=2×n(n+1)2=n(n+1)\sum_{k=1}^{n} 2k = 2 \times \frac{n(n+1)}{2} = n(n+1)

The second group is a geometric sequence:

∑k=0n2k=2n+1−12−1=2n+1−1\sum_{k=0}^{n} 2^k = \frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1

Sn=n(n+1)+2n+1−1S_n = n(n+1) + 2^{n+1} - 1

Answer: Sn=n(n+1)+2n+1−1S_n = n(n+1) + 2^{n+1} - 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
ana_n元素符号a sub nThe nnth term of the original sequence
bn,cnb_n, c_n元素符号b sub n, c sub nThe sub-sequences after grouping

中英对照

中文术语英文术语音标说明
分组求和法grouping method/ˈɡruːpɪŋ ˈmeθəd/The method of summing a sequence by grouping
复合数列composite sequence/ˈkɒmpəzɪt ˈsiːkwəns/A sequence composed of several simple sequences
平方和sum of squares/sʌm əv skweəz/12+22+⋯+n21^2 + 2^2 + \cdots + n^2
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