Quote Originally Posted by Punch View Post
Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$
Here's how to cheat...

If the terms start from $\displaystyle x^0$ upwards, then

$\displaystyle (6-13x+6x^2)(6-13x+6x^2)(6-13x+6x^2)(6-13x+6x^2)(6-13x+6x^2)$

There is a term free of x if we multiply the left terms only..


The lowest power of x is obtained by multiplying the first four 6's, then by -13x...
the middle term in the final factor.


The next highest power of x is found by multiplying the first four 6's by the 3rd term in the last factor.
Then you need to count the number of ways that these calculations can be replicated.
Finally (-13x)(-13x) will also give x "squared" terms.

Of course, it takes too long!