What is the coefficient of x^{3} in this expansion....
(1-x^{2})^{2}(1+x)^{8 }
Thanks!
There is no quick answer unless you use this.
Before we had that, we had to use this: Binomial theorem - Wikipedia, the free encyclopedia
So $\displaystyle (1-x^2)^2 = 1-2x^2+x^4$, and
$\displaystyle (1+x)^8=1+8x+28x^2+56x^3+70x^4+56x^5+28x^6+8x^7+x^ 8$
Picking out the $\displaystyle x^3$ terms gives:
$\displaystyle (1)(56x^3)+(-2x^2)(8x)=56x^3-16x^3=40x^3$, so the coefficient we're looking for is 40.
- Hollywood
By observation we can see that the first expansion will have three terms, one will be constant 1, second with x ^2 and third x^4.
The second expansion will have powers of x from 0 to 8.
Since we have to multiply and get coefficient of x^3, we need to know first two terms of first expansion i.e., 1 -2x^2 and from the second expansion we need to know terms containing x, and x^3 that is 2nd and fourth term.
Thus by knowing four terms we can multiply and get the coefficient of x^3.