# How do I find a coefficient using Pascal's triangle?

Jan 11, 2016

See explanation...

#### Explanation:

The Binomial Theorem for positive integer powers can be written:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

where ((n),(k)) = (n!)/(k! (n-k)!)

Note that some people like to call the first row of Pascal's triangle the $0$th. Others like me prefer to call it the $1$st.

Counting from $1$, the $n + 1$st row of Pascal's triangle consists of the numbers $\left(\begin{matrix}n \\ 0\end{matrix}\right) , \left(\begin{matrix}n \\ 1\end{matrix}\right) , \ldots \left(\begin{matrix}n \\ n\end{matrix}\right)$.

So rather than 'calculate' the individual coefficients for ${\left(a + b\right)}^{n}$, you can read them off from the $\left(n + 1\right)$st row of Pascal's triangle...

For example, if we were calculating ${\left(a + b\right)}^{12}$ then the coefficients would be $1$, $12$, $66$, $220$,..., $1$.

Typically our $a$ and $b$ are not plain variables, but have a multiplier, e.g. ${\left(2 a + 3 b\right)}^{n}$. In such a case you need to multiply the binomial coefficient by a suitable multiple of the powers of $\left(2 a\right)$ and $\left(3 b\right)$, e.g. $\left(\begin{matrix}n \\ k\end{matrix}\right) {\left(2 a\right)}^{n - k} {\left(3 b\right)}^{k} = \left(\begin{matrix}n \\ k\end{matrix}\right) {2}^{n - k} {3}^{k} {a}^{n - k} {b}^{k}$, etc.