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

Jul 31, 2016

${\left(x + 1 + {x}^{- 1}\right)}^{4}$

$= {x}^{4} + 4 {x}^{3} + 10 {x}^{2} + 16 x + 19 + 16 {x}^{- 1} + 10 {x}^{- 2} + 4 {x}^{- 3} + {x}^{- 4}$

#### Explanation:

This is a trinomial, not a binomial. With binomials you could use Pascal's triangle to give you the coefficients.

Since the terms of the given trinomial are in geometric progression, we can use coefficients from a generalisation of Pascal's triangle that adds together $3$ terms rather than $2$:

$\textcolor{w h i t e}{00000000000000} 1$

$\textcolor{w h i t e}{00000000000} 1 \textcolor{w h i t e}{00} 1 \textcolor{w h i t e}{00} 1$

$\textcolor{w h i t e}{00000000} 1 \textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{00} 3 \textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{00} 1$

$\textcolor{w h i t e}{00000} 1 \textcolor{w h i t e}{00} 3 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00} 3 \textcolor{w h i t e}{00} 1$

$\textcolor{w h i t e}{00} 1 \textcolor{w h i t e}{00} 4 \textcolor{w h i t e}{0} 10 \textcolor{w h i t e}{0} 16 \textcolor{w h i t e}{0} 19 \textcolor{w h i t e}{0} 16 \textcolor{w h i t e}{0} 10 \textcolor{w h i t e}{00} 4 \textcolor{w h i t e}{00} 1$

In this triangle each number is the sum of the three numbers above it: left, centre and right.

Hence we find:

${\left(x + 1 + {x}^{- 1}\right)}^{4}$

$= {x}^{4} + 4 {x}^{3} + 10 {x}^{2} + 16 x + 19 + 16 {x}^{- 1} + 10 {x}^{- 2} + 4 {x}^{- 3} + {x}^{- 4}$