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

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- Mar 5th 2009, 10:14 AMalysonbinomial 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 AMGrandadBinomial Coefficients
Hello alyson$\displaystyle \left({\frac{1}{2x}}+x^3\right)^8 = \left(\frac{1+2x^4}{2x}\right)^8$

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

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:

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

Grandad