Question

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

Answer 1

`(y-2)^7 = y^7-14y^6+84y^5-280y^4+560y^3-672y^2+448y-128`

Explanation:

By the binomial theorem:

`(a+b)^n = ((n),(0))a^n + ((n),(1))a^(n-1)b + ((n),(2))a^(n-2)b^2 + ... + ((n),(n))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 `((7),(0)), ((7),(1)), ... , ((7),(7))`

So we find:

`(a+b)^7 = a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^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:

`(y-2)^7 = y^7-14y^6+84y^5-280y^4+560y^3-672y^2+448y-128`