'Binomial Theorem'

Introduction

The binomial theorem provides a formula for expanding (a + b)^n where n is a positive integer, without direct multiplication.

Historical Context

The theorem was known to ancient Indian mathematicians like Pingala (200 BCE). In modern form, it was developed by Isaac Newton and James Gregory.

Binomial Theorem Statement

For any positive integer n: (a + b)^n = C(n,0) a^n b^0 + C(n,1) a^(n-1) b^1 + C(n,2) a^(n-2) b^2 + ... + C(n,n) a^0 b^n

Or in summation notation: (a + b)^n = sum_(k=0)^n C(n,k) a^(n-k) b^k

Understanding the Theorem

Each term C(n,k) a^(n-k) b^k consists of:

• A binomial coefficient C(n,k) = n!/(k!(n-k)!)
• Powers of a and b such that the sum of exponents is always n.

Pascal's Triangle

Each row corresponds to (a + b)^n:

n=0:           1
n=1:         1   1
n=2:       1   2   1
n=3:     1   3   3   1
n=4:    1  4   6   4  1
n=5:  1  5  10  10  5  1

Construction: Each number is the sum of the two numbers directly above it.

Properties:

• First and last entries are always 1.
• The entries are the binomial coefficients C(n, k).
• The triangle is symmetric: C(n, k) = C(n, n-k).

General Term

The (k+1)th term in (a + b)^n is: T_(k+1) = C(n,k) a^(n-k) b^k

This is called the general term because by putting k = 0, 1, 2, ..., n, we get all terms.

Independent Term

A term independent of x (constant term) is found by setting the exponent of x to zero.

Finding a Particular Term

To find the term containing a^p b^q: set n-k = p and k = q, then compute C(n,k).

Middle Term

When n is Even

One middle term: T_(n/2 + 1). Coefficient = C(n, n/2).

When n is Odd

Two middle terms: T_((n+1)/2) and T_((n+3)/2). Coefficients: C(n, (n-1)/2) and C(n, (n+1)/2).

Greatest Coefficient

The greatest binomial coefficient is:

• C(n, n/2) when n is even.
• C(n, (n-1)/2) = C(n, (n+1)/2) when n is odd.

Properties of Binomial Coefficients

In (1 + x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n:

1. C(n,0) + C(n,1) + ... + C(n,n) = 2^n (put x = 1)
2. C(n,0) - C(n,1) + C(n,2) - ... + (-1)^n C(n,n) = 0 (put x = -1)
3. C(n,0) + C(n,2) + C(n,4) + ... = 2^(n-1)
4. C(n,1) + C(n,3) + C(n,5) + ... = 2^(n-1)

Proof of property 1: Put x = 1 in (1 + x)^n = 2^n = sum C(n,k).

Simple Applications

Numerical Approximation

(1.01)^5 = (1 + 0.01)^5 = 1 + 5(0.01) + 10(0.0001) + ... = 1 + 0.05 + 0.001 + ... = 1.05101... (first 3 terms give good approximation)

Divisibility Problems

Prove that 6^n - 5n - 1 is divisible by 25. Write 6^n = (1+5)^n and expand using binomial theorem.

Worked Examples

Example 1: Expand (2x + 3)^4 using binomial theorem. Solution: (2x)^4 + C(4,1)(2x)^3(3) + C(4,2)(2x)^2(3)^2 + C(4,3)(2x)(3)^3 + 3^4 = 16x^4 + 4*8x^3*3 + 6*4x^2*9 + 4*2x*27 + 81 = 16x^4 + 96x^3 + 216x^2 + 216x + 81

Example 2: Find the 5th term in (x + 2y)^10. Solution: T_5 = T_(4+1) = C(10,4) x^(10-4) (2y)^4 = 210 * x^6 * 16y^4 = 3360 x^6 y^4

Example 3: Find the middle term(s) in (x + 1/x)^8. Solution: n = 8 (even). Middle term = T_(8/2+1) = T_5. T_5 = C(8,4) x^(8-4) (1/x)^4 = 70 * x^4 * 1/x^4 = 70

Common Mistakes

1. Incorrect general term: T_(r+1) = C(n,r) a^(n-r) b^r, not C(n,r) a^r b^(n-r).
2. Sign errors in (a-b)^n: (a-b)^n has alternating signs: T_(r+1) = (-1)^r C(n,r) a^(n-r) b^r.
3. Misidentifying middle term: Remember to check if n is even (1 middle term) or odd (2 middle terms).
4. Coefficient vs term: The question often asks for the coefficient, not the full term.

ISC Exam Focus

• Theory (70%): Expansion, general term, middle term, binomial coefficient properties.
• Application (30%): Numerical approximation, divisibility, term independent of x.
• ISC frequently asks: "Find the middle term in the expansion of ...", "Find the coefficient of x^p in ...".
• Typical 4-6 mark questions requiring step-by-step expansion.

Self-Test Questions

Q1: Expand (2 - x)^5. Answer: = 32 - 5*16*x + 10*8*x^2 - 10*4*x^3 + 5*2*x^4 - x^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5

Q2: Find the 6th term in (x + 2y)^9. Answer: T_6 = T_(5+1) = C(9,5) x^(9-5) (2y)^5 = 126 * x^4 * 32 y^5 = 4032 x^4 y^5

Q3: Find the middle term(s) in (2x - y)^7. Answer: n = 7 (odd). Two middle terms: T_4 and T_5. T_4 = C(7,3)(2x)^4(-y)^3 = -35*16x^4 y^3 = -560x^4 y^3. T_5 = C(7,4)(2x)^3(-y)^4 = 35*8x^3 y^4 = 280x^3 y^4.

Q4: Find the term independent of x in (2x + 1/x)^6. Answer: T_(r+1) = C(6,r)(2x)^(6-r)(1/x)^r = C(6,r) 2^(6-r) x^(6-2r). For independent term: 6-2r = 0 => r = 3. So T_4 = C(6,3) 2^3 = 20*8 = 160.

Q5: Prove that C(10,0) + C(10,1) + ... + C(10,10) = 1024. Answer: = 2^10 = 1024.