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Oops, sorry for the typo in the title. :|
Need help with these please:
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:
http://img529.imageshack.us/img529/7942/amtxu3.jpg
Is there a typo in #1. Should it be +63 and not –63?
Ah, crud. You're right. 2 Typos in one post. :(Quote:
Originally Posted by Plato
(1-ax)^n=1 + n(-ax) + n(n-1)(-ax)^2/2 + ..Quote:
Originally Posted by StackFryer
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
rearrange (x-(1/x^2))^8, to:Quote:
Originally Posted by StackFryer
(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