# binomial theorem help

• Mar 5th 2009, 10:14 AM
alyson
binomial theorem help
hey, I'm really struggling on this question i just cant seem to find the right method to get to the answer which is 0.4375

find and simplify the term independent of x in the expansion

$\displaystyle ({\frac{1}{2x}}+x^3)^8$

thanks xx
• Mar 5th 2009, 11:38 AM
Binomial Coefficients
Hello alyson
Quote:

Originally Posted by alyson
hey, I'm really struggling on this question i just cant seem to find the right method to get to the answer which is 0.4375

find and simplify the term independent of x in the expansion

$\displaystyle ({\frac{1}{2x}}+x^3)^8$

thanks xx

$\displaystyle \left({\frac{1}{2x}}+x^3\right)^8 = \left(\frac{1+2x^4}{2x}\right)^8$

$\displaystyle = \left(\frac{1}{2x}\right)^8(1+2x^4)^8$

You now need to find the term in $\displaystyle x^8$ in the expansion of $\displaystyle (1+2x^4)^8$, to cancel the $\displaystyle \frac{1}{x^8}$ outside. This term is:

$\displaystyle \binom{8}{2}(2x^4)^2$

So the coefficient is:

$\displaystyle \frac{1}{2^8}\times \frac{8\times 7 \times 2^4}{2} = \frac{7}{16} = 0.4375$