Hi,
One should recall standard expansions like:
(1) 1 / (1 - x) = 1 + x + x^2 + x^3 + ... + x^n +
...,
of which the coefficient
of x^n is 1.
Note that result (1) follows from the infinite sum of a
geometric progression.
(2) 1 / (1 - x)^2 = 1 + 2x + 3x^2 + 4x^3 + ... + (n +
1)x^n + ...,
of which the coefficient
of x^n is (n + 1).
(3) 1 / (1 - x)^3 = (1/2) [ (1)(2) + (2)(3) x + (3)(4)
x^2 + ... + (n + 1)(n + 2) x^n + ... ],
of which the coefficient
of x^n is (1/2) (n + 1)(n + 2).
In fact, result (2) can be obtained by differentiating
result (1) and result (3) can be reached from result (2) in the
same way.
Differentiation (knowledge from 'O' level) is an easier
approach than the tedious general form involving factorials.
For 1 / (1 + x), 1 / (1 + x)^2 and 1 / (1 + x)^3, we need
to add (-1)^n in front.
P.S. The syllabus does not mention explicitly the coefficient of
the general term. It is also highly unlikely to be asked, since the
last question on it appeared in 1991. The present trend is for
binomial expansion to be asked along with maclaurin's
expansion.
Thanks.
Cheers,
Wen Shih