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Mathematics · Ch 6 — Binomial Theorem

Binomial Theorem for Any Positive Integer n

6.2.1

Binomial Theorem for Any Positive Integer n

The Binomial Theorem for Any Positive Integer nn

The binomial theorem gives us a formula for expanding (a+b)n(a+b)^n when nn is a positive integer — without having to multiply the bracket by itself nn times. The pattern involves binomial coefficients, which you already know as nCr^nC_r (or (nr)\binom{n}{r}).

The theorem states:

(a+b)n= nC0an+ nC1an−1b+ nC2an−2b2+⋯+ nCn−1a bn−1+ nCnbn(a+b)^n = \ ^nC_0 a^n + \ ^nC_1 a^{n-1}b + \ ^nC_2 a^{n-2}b^2 + \dots + \ ^nC_{n-1} a\,b^{n-1} + \ ^nC_n b^n

Each term nCran−rbr^nC_r a^{n-r}b^r has a coefficient nCr^nC_r, and the sum runs from r=0r=0 to r=nr=n. The total number of terms is n+1n+1.

Proof by Mathematical Induction

The theorem is proved using the principle of mathematical induction. Let the statement P(n)P(n) be:

P(n):(a+b)n= nC0an+ nC1an−1b+ nC2an−2b2+⋯+ nCn−1a bn−1+ nCnbnP(n):\quad (a+b)^n = \ ^nC_0 a^n + \ ^nC_1 a^{n-1}b + \ ^nC_2 a^{n-2}b^2 + \dots + \ ^nC_{n-1} a\,b^{n-1} + \ ^nC_n b^n

›Proof

Base case: For n=1n=1, we have

(a+b)1=a+b= 1C0a1+ 1C1b1(a+b)^1 = a+b = \ ^1C_0 a^1 + \ ^1C_1 b^1

Since 1C0=1^1C_0 = 1 and 1C1=1^1C_1 = 1, P(1)P(1) is true.

Induction hypothesis: Assume P(k)P(k) is true for some positive integer kk. That is,

(a+b)k= kC0ak+ kC1ak−1b+ kC2ak−2b2+⋯+ kCkbk…(1)(a+b)^k = \ ^kC_0 a^k + \ ^kC_1 a^{k-1}b + \ ^kC_2 a^{k-2}b^2 + \dots + \ ^kC_k b^k \quad \dots (1)

Induction step: We must prove P(k+1)P(k+1) is true, i.e.,

(a+b)k+1= k+1C0ak+1+ k+1C1akb+ k+1C2ak−1b2+⋯+ k+1Ck+1bk+1(a+b)^{k+1} = \ ^{k+1}C_0 a^{k+1} + \ ^{k+1}C_1 a^k b + \ ^{k+1}C_2 a^{k-1}b^2 + \dots + \ ^{k+1}C_{k+1} b^{k+1}

Start with (a+b)k+1=(a+b)(a+b)k(a+b)^{k+1} = (a+b)(a+b)^k. Using (1):

(a+b)k+1=(a+b)( kC0ak+ kC1ak−1b+ kC2ak−2b2+⋯+ kCk−1a bk−1+ kCkbk)(a+b)^{k+1} = (a+b)\left(\ ^kC_0 a^k + \ ^kC_1 a^{k-1}b + \ ^kC_2 a^{k-2}b^2 + \dots + \ ^kC_{k-1} a\,b^{k-1} + \ ^kC_k b^k\right)

Multiply term-by-term:

= kC0ak+1+ kC1akb+ kC2ak−1b2+⋯+ kCk−1a2bk−1+ kCka bk= \ ^kC_0 a^{k+1} + \ ^kC_1 a^k b + \ ^kC_2 a^{k-1}b^2 + \dots + \ ^kC_{k-1} a^2 b^{k-1} + \ ^kC_k a\,b^k

+ kC0akb+ kC1ak−1b2+ kC2ak−2b3+⋯+ kCk−1a bk+ kCkbk+1+ \ ^kC_0 a^k b + \ ^kC_1 a^{k-1}b^2 + \ ^kC_2 a^{k-2}b^3 + \dots + \ ^kC_{k-1} a\,b^k + \ ^kC_k b^{k+1}

Now group like terms (terms with the same powers of aa and bb):

= kC0ak+1+( kC1+ kC0)akb+( kC2+ kC1)ak−1b2+⋯+( kCk+ kCk−1)a bk+ kCkbk+1= \ ^kC_0 a^{k+1} + \left(\ ^kC_1 + \ ^kC_0\right) a^k b + \left(\ ^kC_2 + \ ^kC_1\right) a^{k-1}b^2 + \dots + \left(\ ^kC_k + \ ^kC_{k-1}\right) a\,b^k + \ ^kC_k b^{k+1}

Use the following identities:

  • k+1C0=1= kC0^{k+1}C_0 = 1 = \ ^kC_0
  • k+1Ck+1=1= kCk^{k+1}C_{k+1} = 1 = \ ^kC_k
  • Pascal's rule:  kCr+ kCr−1= k+1Cr\ ^kC_r + \ ^kC_{r-1} = \ ^{k+1}C_r for 1≤r≤k1 \le r \le k

Applying these:

= k+1C0ak+1+ k+1C1akb+ k+1C2ak−1b2+⋯+ k+1Cka bk+ k+1Ck+1bk+1= \ ^{k+1}C_0 a^{k+1} + \ ^{k+1}C_1 a^k b + \ ^{k+1}C_2 a^{k-1}b^2 + \dots + \ ^{k+1}C_k a\,b^k + \ ^{k+1}C_{k+1} b^{k+1}

This is exactly P(k+1)P(k+1). Hence, whenever P(k)P(k) is true, P(k+1)P(k+1) is also true.

By the principle of mathematical induction, P(n)P(n) is true for every positive integer nn.

A Worked Example: Expanding (x+2)6(x+2)^6

Let's apply the theorem directly:

(x+2)6= 6C0x6+ 6C1x5⋅2+ 6C2x4⋅22+ 6C3x3⋅23+ 6C4x2⋅24+ 6C5x⋅25+ 6C6⋅26(x+2)^6 = \ ^6C_0 x^6 + \ ^6C_1 x^5 \cdot 2 + \ ^6C_2 x^4 \cdot 2^2 + \ ^6C_3 x^3 \cdot 2^3 + \ ^6C_4 x^2 \cdot 2^4 + \ ^6C_5 x \cdot 2^5 + \ ^6C_6 \cdot 2^6

Now compute each binomial coefficient and power of 2:

Term6Cr^6C_r2r2^rProduct
r=0r=011x6x^6
r=1r=16212x512x^5
r=2r=215460x460x^4
r=3r=3208160x3160x^3
r=4r=41516240x2240x^2
r=5r=5632192x192x
r=6r=61646464

So:

(x+2)6=x6+12x5+60x4+160x3+240x2+192x+64(x+2)^6 = x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64

Observations About the Expansion

The textbook lists five important observations. Each one helps you understand the structure of the expansion without having to write it out fully.

1. Sigma Notation for the Binomial Theorem

The expansion can be written compactly using summation notation:

(a+b)n=∑r=0n nCr an−rbr(a+b)^n = \sum_{r=0}^{n} \ ^nC_r \, a^{n-r} b^r

Here, b0=1b^0 = 1 and an−n=a0=1a^{n-n} = a^0 = 1, so the first and last terms are just nC0an^nC_0 a^n and nCnbn^nC_n b^n respectively.

Tip

When writing the sum, remember that rr runs from 0 to nn. The term nCran−rbr^nC_r a^{n-r}b^r is called the general term — it's often denoted Tr+1T_{r+1} because the first term corresponds to r=0r=0.

2. Binomial Coefficients

The numbers nCr^nC_r that appear as coefficients are called binomial coefficients. They are the same as the numbers in Pascal's triangle. For example, in (x+2)6(x+2)^6 above, the coefficients 1,6,15,20,15,6,11, 6, 15, 20, 15, 6, 1 are the binomial coefficients for n=6n=6.

Watch out

Do not confuse binomial coefficients with the numerical coefficients that also include powers of constants. In (x+2)6(x+2)^6, the coefficient of x3x^3 is 160160, but the binomial coefficient is 2020 — the extra factor 88 comes from 232^3.

3. Number of Terms

There are exactly n+1n+1 terms in the expansion of (a+b)n(a+b)^n. This is one more than the exponent nn.

For n=6n=6, we got 7 terms. For n=1n=1, we get 2 terms. This pattern holds for all positive integers nn.

4. Pattern of Exponents

In successive terms: …

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