The Binomial Theorem Extension 1 content

Remember this? Pascal’s Triangle

Some things which may be useful… The sum of the coefficients in any line of Pascal’s Triangle add to 2n.

Jones and Couchman

Using Pascal’s Triangle The expanded form of (1+x)n (where n=1,2,3…) is a polynomial in x with (n+1) terms. The highest power of x is xn. The coefficients are shown by the appropriate line in Pascal’s Triangle.

Generalising Binomial Expansions In the expression (1+x)3 any complex expression can be substituted for x. Hence if (1+x)3 = 1 + 3x + 3x2 + x3 then (1+a4)3 = 1 + 3(a4) + 3(a4)2 + (a4)3 = 1 + 3a4 + 3a8 + a12 Similarly (1- 2𝑥 3 )3 = 1 + 3( −2𝑥 3 ) + 3( −2𝑥 3 )2 + ( −2𝑥 3 )3 = 1 - 2x + 4𝑥 2 3 - 8𝑥 3 27

Jones and Couchman

The expansion of (a+x)n The expanded form of (a+x)n has the same coefficients as (1+x)n. There are (n+1) terms. For any term, the sum of the powers of a and x is n. e.g. Write out the expansion of (𝑥+ 1 𝑥 ) 5 (𝑥+ 1 𝑥 ) 5 = 𝑥 5 + 5𝑥 4 ( 1 𝑥 )+ 10𝑥 3 ( 1 𝑥 ) 2 + 10𝑥 2 ( 1 𝑥 ) 3 + 5𝑥( 1 𝑥 ) 4 +( 1 𝑥 ) 5 e.g. Write out the expansion of (2𝑥 −3𝑦) 4 (2𝑥 −3𝑦) 4 = (2𝑥) 4 + 4( 2𝑥) 3 (−3𝑦)+6( 2𝑥) 2 ( −3𝑦) 2 +4(2𝑥)( −3𝑦) 3 + (−3𝑦) 4 = 16𝑥 4 −96 𝑥 3 𝑦+216 𝑥 2 𝑦 2 −216𝑥 𝑦 3 +81 𝑦 4

Jones and Couchman

The General Binomial Theorem It can be tedious to write out Pascal’s Triangle beyond the first few lines. We need a general formula for binomial expressions. ( 1+𝑥) 𝑛 = 𝑛 0 + 𝑛 1 𝑥+ 𝑛 2 𝑥 2 + 𝑛 3 𝑥 3 +…+ 𝑛 𝑛 𝑥 𝑛 In a similar way: ( 𝑎+𝑥) 𝑛 = 𝑛 0 𝑎 𝑛 + 𝑛 1 𝑎 𝑛−1 𝑥+ 𝑛 2 𝑎 𝑛−2 𝑥 2 + 𝑛 3 𝑎 𝑛−3 𝑥 3 +…+ 𝑛 𝑛 𝑎 0 𝑥 𝑛 We can also write this using sigma notation:

Examples:

Groves

Examples:

Groves

Greatest Coefficient The greatest coefficient can be found without expanding the binomial product:

Examples:

Groves

Proof of Pascal’s Triangle Relations This will be useful in the next proof.

To Prove:

Fitzpatrick