First, there are some results concerning binomial coefficients which will be very useful later.
Worked Examples
Show that \binom{n}{0} = 1, \binom{n}{1} = n and \binom{n}{2} = \frac{n(n - 1)}{2}.
Show me\binom{n}{0} = \frac{n!}{(n-0)!0!} = \frac{n!}{n!0!} = 1 (as 0! = 1)
\binom{n}{1} = \frac{n!}{(n-1)!1!} = \frac{n(n-1)!}{(n-1)!1!} = \frac{n}{1!} = n
\binom{n}{2} = \frac{n!}{(n-2)!2!} = \frac{n(n-1)(n-2)!}{(n-2)!2!} = \frac{n(n-1)}{2!} = \frac{n(n-1)}{2}
Write down corresponding formulae for \binom{n}{3} and \binom{n}{4}.
Show me\binom{n}{3} = \frac{n(n-1)(n-2)}{3!}
\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{4!}
What is \binom{n}{r}?
Show me\binom{n}{r} = \frac{n(n-1)(n-2)...(n-r)}{r!}
The expansion of (1 + x)5 can now be written as follows :
1 + \binom{5}{1}x + \binom{5}{2}x^2 + \binom{5}{3}x^3 + \binom{5}{4}x^4 + x^5
= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5
Similarly the expansion of (a + x)8, where a is any number, can be found thus :
a^8 + \binom{8}{1}a^7x + \binom{8}{2}a^6x^2 + \binom{8}{3}a^5x^3 + \binom{8}{4}a^4x^4 + \binom{8}{5}a^3x^5 + \binom{8}{6}a^2x^6 + \binom{8}{7}ax^7 + x^8
= a^8 + 8a^7x + 28a^6x^2 + 56a^5x^3 + 70a^4x^4 + 56a^3x^5 + 28a^2x^6 + 8ax^7 + x^8
The general result is known as the binomial theorem for a positive integer index.
For a,x \in \R and n \in \N
(a + x)^n = a^n + \binom{n}{1}a^{n-1}x + \binom{n}{2}a^{n - 2}x^2 + ... + \binom{n}{r}a^{n - r}x^r + ... + x^n
Note the way this result is set out. Since n is unknown it is impossible to show the whole expansion. So the formula shows how the expansion starts and finishes, and gives a formula for the general term as
One use of the binomial theorem is to give rough approximations to numbers like (1.01)^8.
You can write
1.01 = 1 + 0.01
so that
(1.01)^8 = 1 + \binom{8}{1}(0.01) + \binom{8}{2}(0.01)^2 + \binom{8}{3}(0.01)^3 + ...
= 1 + 8 × 0.01 + 28 × 0.0001 + 56 × 0.000001 + ...
If 3 s.f. accuracy is required, clearly only the first three terms are needed at most:
(1.01)8 = 1.0828... = 1.08 to 3 s.f.
The first four terms give an accuracy of at least 6 s.f.
| (1.01)8 | = 1 + 0.08 + 0.0028 + 0.000056 |
| = 1.08286 to 6 s.f. |
And all this without a calculator!
One of the really important uses of the Binomial Theorem is with probability where we develop what is called the Binomial Distribution (Unit Probability).
Expand (2x + 3y)4.
| (2 x + 3 y )^4 | = ( 2 x )^4 + \binom{4}{1}(2 x)^3 (3y) + \binom{4}{2}(2 x)^2(3y)^2 + \binom{4}{3}(2 x)(3y)^3 + (3y)^4 |
| = 16 x^4 + 96 x^3 y + 216 x^2 y^2 + 216 xy^3 + 81y^4 |
Expand \left(3-\frac{1}{2}p\right)^5.
| \left(3-\frac{1}{2}p\right)^5 | = 3^5+\binom{5}{1}3^4\left(-\frac{1}{2}p\right)+\binom{5}{2}3^3\left(-\frac{1}{2}p\right)^2+\binom{5}{3}3^2\left(-\frac{1}{2}p\right)^3+\binom{5}{4}3\left(-\frac{1}{2}p\right)^4+\left(-\frac{1}{2}p\right)^5 |
| = 243-\frac{405}{2}p+\frac{135}{2}p^2-\frac{45}{4}p^3+\frac{15}{16}p^4-\frac{1}{32}p^5 |
Find the coefficient of x^5 in the expansion of ( 4 - 3 x )^9.
The x^5 term must be \binom{9}{5}4^4(-3x)^5
= 126 × 256 × (-243)x^5
= -7\ 838\ 208x^5
So the coefficient of x^5 is −7 838 208.
Exercises
