How do you use the Binomial Theorem to expand #(x+1+x^-1)^4#?

1 Answer
Jan 17, 2016

Answer:

See explanation...

Explanation:

This is a multiplication of trinomials not binomials, so the Binomial Theorem does not help much. You could use it by multiplying by #(x-1)^4# then dividing...

The Binomial Theorem allows you to find that:

#(a+b)^4 = sum_(k=0)^4 ((4),(k)) a^(4-k) b^k#

#= a^4+4a^3b+6a^2b^2+4ab^3+b^4#

We can turn our trinomial into a binomial temporarily as follows:

#(x-1)(x+1+x^(-1)) = (x^2-x^(-1))#

Then:

#(x-1)^4(x+1+x^(-1))^4 = (x^2-x^(-1))^4#

#=x^8-4x^5+6x^2-4x^(-1)+x^(-4)#

And:

#(x-1)^4 = x^4-4x^3+6x^2-4x+1#

Then long divide coefficients, including #0#'s for the missing powers of #x# in the dividend...

enter image source here

#(x+1+x^(-1))^4#

#=x^4+4x^3+10x^2+16x+19+16x^(-1)+10x^(-2)+4x^(-3)+x^(-4)#

An easier approach is to simply long multiply coefficients three times:

enter image source here

#(x+1+x^(-1))^4#

#=x^4+4x^3+10x^2+16x+19+16x^(-1)+10x^(-2)+4x^(-3)+x^(-4)#

Trinomial Triangle

Another way of expressing the calculation of these coefficients is to construct a "Trinomial Triangle", similar to Pascal's triangle, but with slightly different rules:

enter image source here

Each number is the sum of the three numbers above it: left, centre and right.

This is not quite as interesting and general in its use as Pascal's triangle, but it does specifically work well for #(x+1+x^(-1))^n#