PracticeVideosBlogFlashcards

14.3 Binomial Theorem, Expansion Terms, and Coefficients

Key Takeaways

  • The Binomial Theorem states (a+b)^n = sum_{r=0}^n C_n^r * a^{n-r} * b^r, where the (r+1)-th term is the general term T_{r+1} = C_n^r * a^{n-r} * b^r.

  • Binomial coefficients C_n^r depend strictly on n and r, being symmetric (C_n^r = C_n^{n-r}) and reaching a maximum at r = n/2 for even n or r = (n-1)/2, (n+1)/2 for odd n.

  • Term coefficients differ from binomial coefficients by including numerical signs and factors from terms a and b.

  • The total sum of binomial coefficients equals 2^n, while the alternating sum equals 0, making odd-indexed and even-indexed coefficient sums each equal to 2^{n-1}.

  • The Value Assignment Method (赋值法) evaluates sums of polynomial expansion coefficients by substituting strategic values such as x = 1, x = 0, or x = -1.

Last updated: July 2026

14.3 Binomial Theorem, Expansion Terms, and Coefficients

The Binomial Theorem establishes an algebraic identity bridging combinatorics, polynomial algebra, and numerical approximation. In Gaokao Mathematics, binomial theorem problems test structural manipulation of general terms, distinction between binomial coefficients and term coefficients, coefficient properties, and value assignment techniques.


1. Statement and Derivation of the Binomial Theorem

1.1 The Binomial Expansion Formula

For any positive integer n∈N∗n \in \mathbb{N}^* and real numbers a,b∈Ra, b \in \mathbb{R}:

(a+b)n=Cn0an+Cn1an−1b+Cn2an−2b2+⋯+Cnran−rbr+⋯+Cnnbn=∑r=0nCnran−rbr(a+b)^n = C_n^0 a^n + C_n^1 a^{n-1}b + C_n^2 a^{n-2}b^2 + \dots + C_n^r a^{n-r}b^r + \dots + C_n^n b^n = \sum_{r=0}^n C_n^r a^{n-r}b^r

Combinatorial Derivation: Consider expanding (a+b)n=(a+b)(a+b)⋯(a+b)(a+b)^n = (a+b)(a+b)\cdots(a+b) (nn factors). Each term in the expanded polynomial is formed by choosing either aa or bb from each of the nn binomial factors. To obtain the term an−rbra^{n-r}b^r, we must choose bb from exactly rr factors and aa from the remaining n−rn-r factors. The number of ways to choose rr factors out of nn to supply bb is CnrC_n^r. Thus, the coefficient of an−rbra^{n-r}b^r is CnrC_n^r.


1.2 The General Term Formula (通项公式)

The (r+1)(r+1)-th term in the expansion of (a+b)n(a+b)^n is designated as the general term Tr+1T_{r+1}:

Tr+1=Cnran−rbr(r=0,1,2,…,n)T_{r+1} = C_n^r a^{n-r} b^r \quad (r = 0, 1, 2, \dots, n)

Important

Critical Distinction in Gaokao Questions:

  1. Index Alignment: Tr+1T_{r+1} represents the (r+1)(r+1)-th term, not the rr-th term. For instance, the 4th term corresponds to r=3r = 3.
  2. Binomial Coefficient vs. Term Coefficient:
    • Binomial Coefficient (二项式系数): Refers strictly to the combinatorial number CnrC_n^r. It depends solely on nn and rr and is always positive.
    • Term Coefficient (项的系数): Refers to the complete numerical multiplier of the algebraic variable in Tr+1T_{r+1}, incorporating signs and constant coefficients present in aa and bb.

2. Fundamental Properties and Coefficient Evaluation Strategies

2.1 Symmetry and Monotonicity of Binomial Coefficients

  1. Symmetry: Cnr=Cnn−rC_n^r = C_n^{n-r}. Terms equidistant from the two ends have equal binomial coefficients.
  2. Maximum Binomial Coefficient (二项式系数的最大值):
    • If nn is even, the middle term r=n2r = \frac{n}{2} has the unique maximum binomial coefficient Cnn/2C_n^{n/2}.
    • If nn is odd, the two central terms r=n−12r = \frac{n-1}{2} and r=n+12r = \frac{n+1}{2} share the maximum binomial coefficient Cn(n−1)/2=Cn(n+1)/2C_n^{(n-1)/2} = C_n^{(n+1)/2}.

2.2 Sum Identities of Binomial Coefficients

By evaluating (1+x)n=∑r=0nCnrxr(1+x)^n = \sum_{r=0}^n C_n^r x^r at specific values of xx:

  1. Total Sum of Binomial Coefficients: Set x=1  ⟹  Cn0+Cn1+Cn2+⋯+Cnn=2nx = 1 \implies C_n^0 + C_n^1 + C_n^2 + \dots + C_n^n = 2^n
  2. Alternating Sum: Set x=−1  ⟹  Cn0−Cn1+Cn2−Cn3+⋯+(−1)nCnn=0x = -1 \implies C_n^0 - C_n^1 + C_n^2 - C_n^3 + \dots + (-1)^n C_n^n = 0
  3. Sums of Even and Odd Power Coefficients: Adding and subtracting the two results yields: Cn0+Cn2+Cn4+⋯=Cn1+Cn3+Cn5+⋯=2n−1C_n^0 + C_n^2 + C_n^4 + \dots = C_n^1 + C_n^3 + C_n^5 + \dots = 2^{n-1}

2.3 The Value Assignment Method (赋值法)

For a polynomial expansion f(x)=(a0+a1x+a2x2+⋯+amxm)n=A0+A1x+A2x2+⋯+Amnxmnf(x) = (a_0 + a_1 x + a_2 x^2 + \dots + a_m x^m)^n = A_0 + A_1 x + A_2 x^2 + \dots + A_{mn} x^{mn}:

  • Constant Term A0A_0: Evaluate f(0)f(0).
  • Sum of All Coefficients ∑Ak\sum A_k: Evaluate f(1)f(1).
  • Alternating Coefficient Sum ∑(−1)kAk\sum (-1)^k A_k: Evaluate f(−1)f(-1).
  • Sum of Even-Indexed Coefficients (A0+A2+A4+…A_0 + A_2 + A_4 + \dots): Calculate f(1)+f(−1)2\frac{f(1) + f(-1)}{2}.
  • Sum of Odd-Indexed Coefficients (A1+A3+A5+…A_1 + A_3 + A_5 + \dots): Calculate f(1)−f(−1)2\frac{f(1) - f(-1)}{2}.
Loading diagram...
Workflow for Binomial Expansion Problems

3. Classic Gaokao Worked Examples and Solutions

Worked Example 1: Finding Specific Coefficients in Rational Power Expansions

Problem Statement: In the expansion of (2x2−1x)9\left(2x^2 - \frac{1}{\sqrt{x}}\right)^9:

  1. Find the term containing x3x^3.
  2. Determine the binomial coefficient and the term coefficient of x3x^3.

Detailed Mathematical Solution:

  • Step 1: Write the general term Tr+1T_{r+1}. Here n=9n = 9, a=2x2a = 2x^2, and b=−x−1/2b = -x^{-1/2}. Tr+1=C9r(2x2)9−r(−x−1/2)rT_{r+1} = C_9^r (2x^2)^{9-r} \left(-x^{-1/2}\right)^r
  • Step 2: Simplify powers of xx and constants. Tr+1=C9r⋅29−r⋅(x2)9−r⋅(−1)r⋅x−r/2T_{r+1} = C_9^r \cdot 2^{9-r} \cdot (x^2)^{9-r} \cdot (-1)^r \cdot x^{-r/2} Tr+1=(−1)r⋅29−r⋅C9r⋅x2(9−r)−r2=(−1)r⋅29−r⋅C9r⋅x18−5r2T_{r+1} = (-1)^r \cdot 2^{9-r} \cdot C_9^r \cdot x^{2(9-r) - \frac{r}{2}} = (-1)^r \cdot 2^{9-r} \cdot C_9^r \cdot x^{18 - \frac{5r}{2}}
  • Step 3: Set the exponent of xx equal to 33. 18−5r2=3  ⟹  5r2=15  ⟹  5r=30  ⟹  r=618 - \frac{5r}{2} = 3 \implies \frac{5r}{2} = 15 \implies 5r = 30 \implies r = 6 Since r=6∈{0,1,…,9}r = 6 \in \{0, 1, \dots, 9\}, the term exists and is T7T_7 (the 7th term).
  • Step 4: Compute Binomial Coefficient and Term Coefficient.
    • Binomial Coefficient: C96=C93=9×8×73×2×1=84C_9^6 = C_9^3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
    • Term Coefficient: Coefficient=(−1)6⋅29−6⋅C96=1⋅23⋅84=8×84=672\text{Coefficient} = (-1)^6 \cdot 2^{9-6} \cdot C_9^6 = 1 \cdot 2^3 \cdot 84 = 8 \times 84 = 672
    • The term is 672x3672 x^3.

Worked Example 2: Product Expansion Coefficient Calculation

Problem Statement: Find the coefficient of x4x^4 in the expansion of (1+2x)5(1−x)4(1 + 2x)^5 (1 - x)^4.

Detailed Mathematical Solution:

  • Step 1: Express individual expansions.

    • Expansion of (1+2x)5(1+2x)^5: General term Ar+1=C5r(2x)r=C5r⋅2r⋅xrA_{r+1} = C_5^r (2x)^r = C_5^r \cdot 2^r \cdot x^r for r∈{0,1,2,3,4,5}r \in \{0, 1, 2, 3, 4, 5\}.
    • Expansion of (1−x)4(1-x)^4: General term Bk+1=C4k(−x)k=C4k(−1)kxkB_{k+1} = C_4^k (-x)^k = C_4^k (-1)^k x^k for k∈{0,1,2,3,4}k \in \{0, 1, 2, 3, 4\}.
  • Step 2: Form the combined power x4x^4. The product of a term with xrx^r and a term with xkx^k gives xr+kx^{r+k}. We require r+k=4r + k = 4 where 0≤r≤50 \le r \le 5 and 0≤k≤40 \le k \le 4. We list all non-negative integer pairs (r,k)(r, k):

    1. Pair (4,0)(4, 0): r=4,k=0r = 4, k = 0
    2. Pair (3,1)(3, 1): r=3,k=1r = 3, k = 1
    3. Pair (2,2)(2, 2): r=2,k=2r = 2, k = 2
    4. Pair (1,3)(1, 3): r=1,k=3r = 1, k = 3
    5. Pair (0,4)(0, 4): r=0,k=4r = 0, k = 4
  • Step 3: Calculate product terms for each pair.

    • Pair (4,0)(4, 0): [C5424]×[C40(−1)0]=(5×16)×1=80\left[C_5^4 2^4\right] \times \left[C_4^0 (-1)^0\right] = (5 \times 16) \times 1 = 80
    • Pair (3,1)(3, 1): [C5323]×[C41(−1)1]=(10×8)×(4×−1)=80×(−4)=−320\left[C_5^3 2^3\right] \times \left[C_4^1 (-1)^1\right] = (10 \times 8) \times (4 \times -1) = 80 \times (-4) = -320
    • Pair (2,2)(2, 2): [C5222]×[C42(−1)2]=(10×4)×(6×1)=40×6=240\left[C_5^2 2^2\right] \times \left[C_4^2 (-1)^2\right] = (10 \times 4) \times (6 \times 1) = 40 \times 6 = 240
    • Pair (1,3)(1, 3): [C5121]×[C43(−1)3]=(5×2)×(4×−1)=10×(−4)=−40\left[C_5^1 2^1\right] \times \left[C_4^3 (-1)^3\right] = (5 \times 2) \times (4 \times -1) = 10 \times (-4) = -40
    • Pair (0,4)(0, 4): [C5020]×[C44(−1)4]=1×1=1\left[C_5^0 2^0\right] \times \left[C_4^4 (-1)^4\right] = 1 \times 1 = 1
  • Step 4: Sum all partial coefficients.

    Total Coefficient of x4=80−320+240−40+1=−39\text{Total Coefficient of } x^4 = 80 - 320 + 240 - 40 + 1 = -39
Test Your Knowledge

What is the coefficient of x^3 in the expansion of (x - 2/x)^7?

A

-84

B

84

C

-280

D

280

Test Your Knowledge

If (1+2x)6=a0+a1x+a2x2+⋯+a6x6(1 + 2x)^6 = a_0 + a_1 x + a_2 x^2 + \dots + a_6 x^6, what is the value of the coefficient sum a1+a2+a3+a4+a5+a6a_1 + a_2 + a_3 + a_4 + a_5 + a_6?

A

0

B

729

C

728

D

364

Test Your Knowledge

Find the constant term (the term independent of x) in the expansion of (x^2 + 2/x)^6.

A

60

B

120

C

180

D

240

Sections you finish are checked off in the contents.