Binomial Theorem Help PLZ! (1 Viewer)

[aanthnnyyy]

Find coefficient of x*2 in (x+1/x)^3•(x+1)^5

I know it's easy when there is one bracket... But when they double up how do we do it?

 

[leehuan]

I'm guessing you meant coefficient of x^2

We consider cases seperately:
We can get x^2 by either:
- Taking x^(-3) from the left bracket and multiplying by x^5 from the right bracket
- Taking x^(-2) from the left bracket and multiplying by x^4 from the right bracket
- Taking x^(-1) from left and x^3 from right
- Taking constants from the left and x^2 from the right
- Taking x from the left and x from the right
- Taking x^2 from the left and constants from the right

So the coefficient of x^2 turns out to be:
(Coeff of x^(-3) from left)*(Coeff of x^5 from right) + (Coeff of x^(-2) from left)*(Coeff of x^4 from right) + ... + (Coeff of x^3 from left)*(Constants from right)

 

[integral95]

The powers of the 4 terms in the first bracket when expanded is

To get the coefficients of x^2 it's

 

[aanthnnyyy]

Ahhhh Thanks. I know I had to match up powers but this makes so much more sense. I find this harder then proving via differentiation a integration :s