Understanding Engineering Mathematics

The expansion for( 1 +x)nis now easily obtained from the above by puttinga=1,

b=xto get

( 1 +x)n= 1 +nx+

n(n− 1 )

2!

x^2 +

n(n− 1 )(n− 2 )

3!

x^3 +···+xn

= 1 +nC 1 x+nC 2 x^2 +nC 3 x^3 +···+xn

You may have previously seen the binomial theorem treated by Pascal’s triangle. While

this is fine for expanding small powers such as( 1 +x)^4 , it is cumbersome for increasingly

higher powers and less useful for more advanced theory and use of the binomial theorem,

as for example in determining a required coefficient in the expansion. Note also that while

here we have takennto be a positive integer, it is possible to extend the theorem to

the cases ofnnegative or a fraction. In such cases the above series expansion never

terminates (➤106).

Solution to review question 2.1.13

The sign needs care here:

( 2 −x)^5 = 25 + 5 × 24 (−x)+

5 × 4

2!

23 (−x)^2

+

5 × 4 × 3

3!

22 (−x)^3 +

5 × 4 × 3 × 2

4!

2 (−x)^4

+

5 × 4 × 3 × 2 × 1

5!

(−x)^5

= 32 − 5 × 16 x+ 10 × 8 x^2 − 10 × 4 x^3 + 10 x^4 −x^5

= 32 − 80 x+ 80 x^2 − 40 x^3 + 10 x^4 −x^5

You may have obtained the coefficients in a different way – say by Pascal’s

triangle, or, you may have factored out the 2 and used the expansions of

( 1 +x)ninstead of arriving directly at the penultimate line. In any event

a steady hand is required to avoid losing signs or factors of 2. Finally

notice a simple check for your result, by puttingx=1 in both sides.

Whenx=1,( 2 −x)^5 = 15 =1, while the expanded result gives

32 − 80 + 80 − 40 + 10 − 1 = 1

which checks.

2.3 Reinforcement

2.3.1 Multiplication of linear expressions

➤➤

38 40

➤

A.Identify which of the following are algebraic expressions in the variables involved:

(i) 14 (ii) x− 3

(iii) 2t+ 1 =0(iv)2x+y