Binomial Expansion Question

Okay, another question.

The first four terms in the series expansion of (1+ax)^n in ascending powers of x are

1 - 4x + 24x^2 + kx^3

where a, n and k are constants and |ax| < 1.

(i) Find the values of a and n.

(ii) Show that k = -160.

I got a = -1 and n = 4 for the first part, but they don't work for part two. :confused:

Reply 1

Twiglet

heya

If you write out (1+ax)^n as 1+ nax + (n(n-1)a^2)/2.x^2 + (n(n-1)(n-2)a^3)/6.x^3

so taking x: na=-4, so n=-4/a (1)
and taking x^2: na^2(n-1)/2 = 24 (2)

so sub 1 into 2
-4/a.a^2 (-4/a -1)=48
16+ 4a = 48
a=8
therefore n=-0.5

using k= (n(n-1)(n-2)a^3)/6 = (-0.5 x -1.5 x -2.5)/6 x 8^3 = -160 :wink: