1. ## Pascal's Trianglw

Oops, sorry for the typo in the title. :|

In the expansion of (1-ax)^n, the first three terms are 1-12x-63x^2. Find the values of a and n.

and

In the expansion of (x-(1/x^2))^8 find:

a) The term that has x^-1

b) The coefficient of the term that has x^2.

Just to make the second one a little clearer, here's something I whipped up with my MSPaint skills:

2. Is there a typo in #1. Should it be +63 and not –63?

3. Originally Posted by Plato
Is there a typo in #1. Should it be +63 and not –63?
Ah, crud. You're right. 2 Typos in one post.

4. Originally Posted by StackFryer
Oops, sorry for the typo in the title. :|

In the expansion of (1-ax)^n, the first three terms are 1-12x+63x^2. Find the values of a and n.
(1-ax)^n=1 + n(-ax) + n(n-1)(-ax)^2/2 + ..

so:

a.n=12

and:

(n(n-1)a^2)/2=63.

Expanding this last equation:

(na)^2-(na)a=126,

or:

144-12a=126

so:

a=3/2

hence:

n=8.

RonL

5. Originally Posted by StackFryer
In the expansion of (x-(1/x^2))^8 find:

a) The term that has x^-1

b) The coefficient of the term that has x^2.

Just to make the second one a little clearer, here's something I whipped up with my MSPaint skills:
rearrange (x-(1/x^2))^8, to:

(1/x^16) (1-x^3)^8 = (1/x^16)[1+8x^3+28x^6+54x^9+70x^12+54x^15+

...................28x^18+8x^21+x^24]

a) so the coefficient of x^(-1) in this is the coefficient of x^15 in the []'s,
so is 54.

b) so the coefficient of x^2 in this is the coefficient of x^18 in the []'s,
so is 28.

RonL