Binomial Theorem

Printable View

March 21st 2010, 11:33 AM
DemonX01

Binomial Theorem

Given that

(1+x/2)^8(1-x)^6)≡1+Ax+Bx^2...

find the constants A and B

how would I solve for them, other than completely expanding each binomial?
March 21st 2010, 12:37 PM
Plato

Quote:

Originally Posted by DemonX01

(1+x/2)^8(1-x)^6)≡1+Ax+Bx^2...

find the constants A and B

$\left(1+\frac{x}{2}\right)^8=1+4x+14x^2+\cdots$
$(1-x)^6=1-x+15x^2+\cdots$

Mulriply those and see what happens.

All times are GMT -8. The time now is 06:15 PM.

Copyright © 2005-2014 Math Help Forum. All rights reserved.

Copyright © 2005-2014 Math Help Forum. All rights reserved.