Home > Algebra, Number Theory > Binomial Theorem (Expansion) for positive integer powers

Binomial Theorem (Expansion) for positive integer powers

August 17th, 2010

The binomial theorem (also called binomial expansion) allows us to find the value of a sum of two items raised to any given power, in this post we will consider the simpler example when that power is a positive integer. So in particular we will consider the expansion of (a+b)n.

When multiply out this bracket we get all the combinations of powers for a and b such that the sum of their powers is n. So there is a term of anwhere the power of b is 0 and one of bn and then there are terms in between with both a and b raised to powers. Then all that is left is to find the co-efficents of each of these terms. For the case that n is a positive integer there are two ways of doing this, one is to use the Binomial Theorem or the other is to use Pascal’s triangle.

Binomial Theorem

The binomial Theorem is:
(1+x)^n = \sum^n_{i=0}{\binom{n}{i} x^i} = \sum^n_{i=0}{\frac{n!}{i!(n-i)!}x^i}

Example
(1+3)^4 = \frac{4.3.2.1.3^0}{4.3.2.1} + \frac{4.3.2.1.3^1}{(3.2.1)(1)} + \frac{4.3.2.1.3^2}{(2.1)(2.1)} + \frac{4.3.2.1.3^3}{(1)(3.2.1)} + \frac{4.3.2.1.3^4}{4.3.2.1}

If we know calculate the factorials this becomes

(1+3)^4 = 1 + 4.3 + 6.3^2 + 4.3^3 + 3^4 = 256

Which you can check since we know 3+1 = 4 and 44 = 256.

However there is another simpler way of finding the coefficients for the power which is to use a Pascal’s triangle. This is an quick to calculate way for moderately large powers (the triangle can get quite large for big powers above say 10) by hand. A Pascal’s triangle with 5 rows is below.

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Each item in the triangle is the sum of the two numbers above it and there is a 1 at the end of each row. When the binomial terms are written in order of powers of x (either way as it is symmetric) each number in the nth row is the coefficient of x in corresponding term (check it with the example above)

More complex sums

We can also use the binomial theorem on sums where the first term is something other than 1. In general when we want to find (a+b)n.
To do this we first factor out a within the sum to get
(a + b)^n = (a(1 + \frac{b}{a}))^n = a^n \cdot (1 + \frac{b}{a})^n
We then treat \frac{b}{a} as x and use the binomial theorem to expand (1+\frac{b}{a})^n to get

(a + b)^n = a^n [\sum^n_{i=0}{\binom{n}{i}\frac{b}{a}^i}]

we can then multiply through the sum by an to get

(a + b)^n = \sum^n_{i=0}{\binom{n}{i}\frac{b^i}{a^i}a^n} = \sum^n_{i=0}{\binom{n}{i}b^{i}a^{n-i} }

Example

Expand (3 + 2x)4.

Firstly factorise the 3 to get
(3+2x)^4 = 3^4 (1 + \frac{2x}{3})^4)
Using Pascal’s triangle or our previous calculations we can find the coefficients and write out the sum
(3 + 2x)^4 = 3^4 + 4 \cdot 3^3\cdot + 6\cdot3^2\cdot (2x)^2 + 4\cdot \cdot (2x)^3 + (2x)^4
and now working out the arithmetic of each term we get our final answer
(3+2x)^4 = 81 + 216x + 216x^2 + 96x^3 + 16x^4

The binomial theorem can also be used to calculated sums raised to non integer or negative powers but that is for another post.

Categories: Algebra, Number Theory Tags: , , , ,