Understanding Engineering Mathematics

but this soon becomes tedious. Fortunately there is a well established routine method for
expanding such expressions. Sincea+bis abinomialthis method is called thebinomial
theorem.
The binomial theorem deals with expanding expressions such as(a+b)nor( 1 +x)n.
Here we will restrictnto be a positive integer, but it works equally well for other values
(leading to infinite series). This is perhaps a place where ‘in at the deep end’ is the best
policy and so here it is, the binomial theorem:

(a+b)n=an+nan−^1 b+

n(n− 1 )

2!

an−^2 b^2 +···

+

n(n− 1 )...(n−r+ 1 )

r!

an−rbr+···+bn

or, using the combinatorial notationnCr(Section 1.2.6➤)

(a+b)n=an+nC 1 an−^1 b+nC 2 an−^2 b^2 +···

+nCran−rbr+···+bn

You have two problems (at least!) here – remembering the result and understanding where
it comes from. Memorising it is helped by noting the patterns:

• terms are of the forman−rbr,i.e.nquantities multiplied together


• the coefficient ofan−rbrisnCr=

n!

(n−r)!r!

=

n(n− 1 )...(n−r+ 1 )

r!

• there is symmetry – for example terms of the forman−rbrandarbn−r
  are symmetrically placed, as are the coefficients, sincenCr=nCn−r


• the coefficient of an−rbr has r! in the denominator and
  n(n− 1 )...(n−r+ 1 )in the numerator. The latter is perhaps best
  remembered by starting at the second term withnand reducingnin
  successive terms.

Example

(a+b)^6 =a^6 + 6 a^5 b+

6. 5

2!

a^4 b^2 +

6. 5. 4

3!

a^3 b^3

+

6. 5. 4. 3

4!

a^2 b^4 +

6. 5. 4. 3. 2

5!

ab^5 +

6!

6!

b^6

=a^6 + 6 a^5 b+ 15 a^4 b^2 + 20 a^3 b^3 + 15 a^2 b^4 + 6 ab^5 +b^6

Your second problem, understanding where the binomial expansion comes from, is best
approached by noticing that in

(a+b)n=(a+b)×(a+b)×···×(a+b)

︸ ︷︷ ︸

nfactors

we can form terms of the forman−rbrby choosing, say,binrways fromndifferent
brackets – there arenCr ways to do this, givingnCr such terms in all, contributing
nC
ra
n−rbrto the expansion.