Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
We concern ourselves with expanding powers of the form
(a + b)n , where a and b are algebraic expressions (usually
variables or real number constants) and n is a nonnegative
integer.
We have previously observed
(a + b)0 = 1.
(a + b)1 = a + b.
(a + b)2 = a2 + 2ab + b 2 .
(a + b)3 = a3 + 3a2 b + 3ab 2 + b 3 .
(a + b) = (a + b)3 (a + b) = a4 + 4ab 2 + 6a2 b 2 + 4a3 b + b 4 .
4
, Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
Pascal’s triangle
The coefficients of (a + b)n may be computed from the coefficients
of (a + b)n−1 according to what is known as Pascal’s triangle, the
first five rows of which are displayed below:
Pascal’s triangle
(a + b)0 1
(a + b)1 1 1
(a + b)2 1 2 1
(a + b)3 1 3 3 1
(a + b)4 1 4 6 4 1
(a + b)5 1 5 10 10 5 1
, Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
Example 2.19 Use Pascal’s triangle to determine the following
expansions:
(i) (1 + x)6 .
(ii) (2x − y )4 .
, Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
Solutions:
(i) Here, a = 1 and b = x. The sixth row of Pascal’s triangle is
1 6 15 20 15 6 1, so
(1 + x)6 = 1 · 16 x 0 + 6 · 15 x 1 + 15 · 14 x 2 +
+20 · 13 x 3 + 15 · 12 x 4 + 6 · 11 x 5 + 1 · 10 x 6 =
= 1 + 6x + 15x 2 + 20x 3 + 15x 4 + 6x 5 + x 6 .
Partial fractions The binomial theorem
We concern ourselves with expanding powers of the form
(a + b)n , where a and b are algebraic expressions (usually
variables or real number constants) and n is a nonnegative
integer.
We have previously observed
(a + b)0 = 1.
(a + b)1 = a + b.
(a + b)2 = a2 + 2ab + b 2 .
(a + b)3 = a3 + 3a2 b + 3ab 2 + b 3 .
(a + b) = (a + b)3 (a + b) = a4 + 4ab 2 + 6a2 b 2 + 4a3 b + b 4 .
4
, Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
Pascal’s triangle
The coefficients of (a + b)n may be computed from the coefficients
of (a + b)n−1 according to what is known as Pascal’s triangle, the
first five rows of which are displayed below:
Pascal’s triangle
(a + b)0 1
(a + b)1 1 1
(a + b)2 1 2 1
(a + b)3 1 3 3 1
(a + b)4 1 4 6 4 1
(a + b)5 1 5 10 10 5 1
, Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
Example 2.19 Use Pascal’s triangle to determine the following
expansions:
(i) (1 + x)6 .
(ii) (2x − y )4 .
, Binomial expansions and the Binomial Theorem Binomial coefficients
Partial fractions The binomial theorem
Solutions:
(i) Here, a = 1 and b = x. The sixth row of Pascal’s triangle is
1 6 15 20 15 6 1, so
(1 + x)6 = 1 · 16 x 0 + 6 · 15 x 1 + 15 · 14 x 2 +
+20 · 13 x 3 + 15 · 12 x 4 + 6 · 11 x 5 + 1 · 10 x 6 =
= 1 + 6x + 15x 2 + 20x 3 + 15x 4 + 6x 5 + x 6 .
