# How do I find the constant term of a binomial expansion?

Sep 26, 2014

The constant term of ${\left(x + a\right)}^{n}$ is always ${a}^{n}$; for example, the constant term of ${\left(x + 3\right)}^{7}$ is ${3}^{7}$.

Jun 3, 2015

The expansion of a binomial is given by the Binomial Theorem:

${\left(x + y\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) \cdot {x}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) \cdot {x}^{n - 1} \cdot {y}^{1} + \ldots + \left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {y}^{k} + \ldots + \left(\begin{matrix}n \\ n\end{matrix}\right) \cdot {y}^{n} = {\sum}_{k = 0}^{n} \cdot \left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {y}^{k}$
where $x , y \in \mathbb{R}$, $k , n \in \mathbb{N}$, and $\left(\begin{matrix}n \\ k\end{matrix}\right)$ denotes combinations of $n$ things taken $k$ at a time.

$\left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {y}^{k}$ is the general term of the binomial expansion.

We also have the formula: ( (n), (k) )=(n!)/(k!*(n-k)!), where k! = 1*2*...*k

We have three cases:

Case 1: If the terms of the binomial are a variable and a constant (y=c, where $c$ is a constant), we have ${\left(x + c\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) \cdot {x}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) \cdot {x}^{n - 1} \cdot {c}^{1} + \ldots + \left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {c}^{k} + \ldots + \left(\begin{matrix}n \\ n\end{matrix}\right) \cdot {c}^{n}$

We can see that the constant term is the last one: $\left(\begin{matrix}n \\ n\end{matrix}\right) \cdot {c}^{n}$
(as $\left(\begin{matrix}n \\ n\end{matrix}\right)$ and ${c}^{n}$ are constant, their product is also a constant).

Case 2: If the terms of the binomial are a variable and a ratio of that variable ($y = \frac{c}{x}$, where $c$ is a constant), we have:
${\left(x + \frac{c}{x}\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) \cdot {x}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) \cdot {x}^{n - 1} \cdot {\left(\frac{c}{x}\right)}^{1} + \ldots + \left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {\left(\frac{c}{x}\right)}^{k} + \ldots + \left(\begin{matrix}n \\ n\end{matrix}\right) \cdot {\left(\frac{c}{x}\right)}^{n}$

This time, we see that the constant term is not to be found at the extremities of the binomial expansion. So, we should have a look at the general term and try to find out when it becomes a constant:
$\left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {\left(\frac{c}{x}\right)}^{k} = \left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {x}^{n - k} \cdot {c}^{k} \cdot \frac{1}{x} ^ k = \left(\left(\begin{matrix}n \\ n\end{matrix}\right) \cdot {c}^{k}\right) \cdot \frac{{x}^{n - k}}{x} ^ k = \left(\left(\begin{matrix}n \\ k\end{matrix}\right) \cdot {c}^{k}\right) \cdot {x}^{n - 2 k}$.

We can see that the general term becomes constant when the exponent of variable $x$ is $0$. Therefore, the condition for the constant term is: $n - 2 k = 0 \Rightarrow$ $k = \frac{n}{2}$ . In other words, in this case, the constant term is the middle one ($k = \frac{n}{2}$).

Case 3: If the terms of the binomial are two distinct variables $x$ and $y$, such that $y$ cannot be expressed as a ratio of $x$, then there is no constant term . This is the general case ${\left(x + y\right)}^{n}$