Question

How do you expand `(r-t)^12` using the binomial theorem?

Answer 1

`r^12-12r^11t+66r^10t^2-220r^9t^3+495r^8t^4-792r^7t^5+924r^6t^t-792r^5t^7+495r^4t^8-220r^3t^9+66r^2t^10-12rt^11+t^12`

Explanation:

The binomial theorem tells us that:

`(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k`

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

The binomial coefficients `((n),(k))` can be read from the appropriate row of Pascal's triangle.

In our example, with `n=12` we need the row beginning `1, 12`.

Putting `a=r` and `b=-t`, we also find that we need to alternate the signs of consecutive terms.

Hence we find:

`(r-t)^12`

`=r^12-12r^11t+66r^10t^2-220r^9t^3+495r^8t^4-792r^7t^5+924r^6t^t-792r^5t^7+495r^4t^8-220r^3t^9+66r^2t^10-12rt^11+t^12`