Chapter 7 Solutions — Binomial Theorem

Using the Binomial Theorem: (a + b)^n = Σ C(n,r) a^(n-r) b^r

Here a = 1, b = –2x, n = 5.

(1 – 2x)⁵ = C(5,0)(1)⁵ + C(5,1)(1)⁴(–2x) + C(5,2)(1)³(–2x)² + C(5,3)(1)²(–2x)³ + C(5,4)(1)(–2x)⁴ + C(5,5)(–2x)⁵

= 1 + 5(–2x) + 10(4x²) + 10(–8x³) + 5(16x⁴) + (–32x⁵)

= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵

Write a = (a – b) + b, so a^n = ((a – b) + b)^n.

By the Binomial Theorem:

a^n = Σ (r=0 to n) C(n,r) (a – b)^r · b^(n-r)

= C(n,0)·b^n + C(n,1)(a–b)·b^(n-1) + C(n,2)(a–b)²·b^(n-2) + … + (a–b)^n

= b^n + (a–b)[C(n,1)·b^(n-1) + C(n,2)(a–b)·b^(n-2) + … + (a–b)^(n-1)]

Therefore:

a^n – b^n = (a – b)[C(n,1)·b^(n-1) + C(n,2)(a–b)·b^(n-2) + … + (a–b)^(n-1)]

The expression in brackets is an integer (since a, b are integers and n is a positive integer).

Hence (a – b) is a factor of a^n – b^n. (Proved)

Here a = 2/x, b = –x/2, n = 5.

(2/x – x/2)⁵ = C(5,0)(2/x)⁵ + C(5,1)(2/x)⁴(–x/2) + C(5,2)(2/x)³(–x/2)² + C(5,3)(2/x)²(–x/2)³ + C(5,4)(2/x)(–x/2)⁴ + C(5,5)(–x/2)⁵

Term 1: (2/x)⁵ = 32/x⁵

Term 2: 5 · (16/x⁴) · (–x/2) = 5 · (–8/x³) = –40/x³

Term 3: 10 · (8/x³) · (x²/4) = 10 · (2/x) = 20/x

Term 4: 10 · (4/x²) · (–x³/8) = 10 · (–x/2) = –5x

Term 5: 5 · (2/x) · (x⁴/16) = 5 · (x³/8) = 5x³/8

Term 6: (–x/2)⁵ = –x⁵/32

= 32/x⁵ – 40/x³ + 20/x – 5x + (5x³)/8 – x⁵/32