# How can you use the binomial theorem to expand (y-2)^7 ?

Sep 15, 2016

${\left(y - 2\right)}^{7} = {y}^{7} - 14 {y}^{6} + 84 {y}^{5} - 280 {y}^{4} + 560 {y}^{3} - 672 {y}^{2} + 448 y - 128$

#### Explanation:

By the binomial theorem:

${\left(a + b\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} b + \left(\begin{matrix}n \\ 2\end{matrix}\right) {a}^{n - 2} {b}^{2} + \ldots + \left(\begin{matrix}n \\ n\end{matrix}\right) {b}^{n}$

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

Rather than mess with all those factorials directly, we can pick out the appropriate row of Pascal's triangle, which in our example is the row beginning $1 , 7$. This row of Pascal's triangle gives us the values $\left(\begin{matrix}7 \\ 0\end{matrix}\right) , \left(\begin{matrix}7 \\ 1\end{matrix}\right) , \ldots , \left(\begin{matrix}7 \\ 7\end{matrix}\right)$

So we find:

${\left(a + b\right)}^{7} = {a}^{7} + 7 {a}^{6} b + 21 {a}^{5} {b}^{2} + 35 {a}^{4} {b}^{3} + 35 {a}^{3} {b}^{4} + 21 {a}^{2} {b}^{5} + 7 a {b}^{6} + {b}^{7}$

In our example, $a = y$ and $b = - 2$, so we have some powers of $2$ and alternating signs to deal with.

The easiest way is probably to write out some sequences to construct our coefficients...

Here's the row from Pascal's triangle:

$1 , 7 , 21 , 35 , 35 , 21 , 7 , 1$

Here are powers of $2$ in ascending order:

$1 , 2 , 4 , 8 , 16 , 32 , 64 , 128$

Multiply the two sequences together to get:

$1 , 14 , 84 , 280 , 560 , 672 , 448 , 128$

Then alternate the signs:

$1 , - 14 , 84 , - 280 , 560 , - 672 , 448 , - 128$

These are the coefficients we need. So we have:

${\left(y - 2\right)}^{7} = {y}^{7} - 14 {y}^{6} + 84 {y}^{5} - 280 {y}^{4} + 560 {y}^{3} - 672 {y}^{2} + 448 y - 128$