Question

How do you expand `(r+s)^8`?

Answer 1

`(r+s)^8 = r^8+8r^7s+28r^6s^2+56r^5s^3+70r^4s^4+56r^3s^5+28r^2s^6+8rs^7+s^8`

Explanation:

By the binomial theorem:

`(r+s)^8 = sum_(k=0)^8 ((8),(k)) r^(8-k) s^k`

where `((8),(k)) = (8!)/((8-k)! k!)`

Rather than calculate these binomial coefficients explicitly, we can read them from the `9`th row of Pascal's triangle (i.e. the row beginning `1, 8,...`)

So:

`(r+s)^8 = r^8+8r^7s+28r^6s^2+56r^5s^3+70r^4s^4+56r^3s^5+28r^2s^6+8rs^7+s^8`