The Binomial Series

Pascal’s Triangle

You should know that (a + b)² = a² + 2ab + b² and you should be able to work out that (a + b)³ = a³ + 3a²b + 3b²a + b³ .
It should also be obvious to you that (a + b)¹ = a + b .

so (a + b)¹  =        a + b
(a + b)² =       a² + 2ab + b²
(a + b)³ =  a³ + 3a²b + 3b²a + b³

You should notice that the coefficients of (the numbers before) a and b are:
     1   1
  1   2   1
1   3   3   1

If you continued expanding the brackets for higher powers, you would find that the sequence continues:
      1   4   6   4   1
   1   5  10  10  5   1
1   6  15  20  15  6   1
etc

This sequence is known as Pascal's triangle. Each of the numbers is found by adding together the two numbers directly above it.
So the 20 in the last line is found by adding together 10 and 10. Each of the 10s in the line above are found by adding together a 6 and a 4.

So it is possible to expand (a + b) to any whole number power by knowing Pascal's triangle.

Example

Find (3 + x)³

The power that we are expanding the bracket to is 3, so we look at the third line of Pascal’s triangle, which is 1 3 3 1.

So the answer is: 3³ + 3 × (3² × x) + 3 × (x² × 3) + x³(we are replacing a by 3 and b by x in the expansion of (a + b)³ above)

Generally

It is, of course, often impractical to write out Pascal's triangle every time, when all that we need to know are the entries on the nth line. Clearly, the first number on the nth line is 1. The second number is n. The third number is:
n(n - 1)   .
 1 × 2

In general, the rth number in the nth line is:
    n!        (which is ⁿC_r on your calculator)
r! (n - r)!

where n! means ‘n factorial’ and is equal to n × (n-1) × … × 2 × 1

ⁿC_r is also often written as and is pronounced “n choose r”.

The Binomial Theorem

The Binomial Theorem states that, where n is a positive integer:

• (a + b)ⁿ = aⁿ + (ⁿC₁)a^n-1b + (ⁿC₂)a^n-2b² + … + (ⁿC_n-1)ab^n-1 + bⁿ

Example

Expand (4 + 2x)⁶ in ascending powers of x up to the term in x³

This means use the Binomial theorem to expand the terms in the brackets, but only go as high as x³.

So to find the answer we substitute 4 for a in the Binomial theorem and 2x for b:

4⁶ + (⁶C₁)(4⁵)(2x) + (⁶C₂)(4⁴)(2x)² + (⁶C₃)(4³)(2x)³ + …
= 4096 + (6 ×1024 ×2x) + (15 ×256 ×4x²) + (20 ×64 ×8x³) + …
= 4096 + 12288x + 15360x² + 10240x³ + …