The middle term in the expansion of (1/x + sqr root of x)^8 is?

Question

anonymous · Answer

do you mean this??
$$\Large \left(\frac1x+\sqrt x\right)^8$$

callisto · Answer

in the expansion of (1/x + sqr root of x)^8 , there are 9 terms. So the middle term is the 5th term
5th term  $$C_{5-1}^{8}(1/x)^{8-(5-1)}\sqrt{x}^{5-1} $$$$= C_{4}^{8}(1/x)^{4}\sqrt{x}^{4}$$$$=70 [(x^2)/(x^{4})$$
Can you do it from here?

anonymous · Answer

$$\dbinom{8}{4}(\frac{1}{x})^4\sqrt{x}^4$$ is a start

didee · Answer

@Kreshnik, yes that is the question

@Callisto, that seems right. Please post the original formula cause I have more of these type of questions, thanks

callisto · Answer

the general rth term of expansion of (a+b)^n
$$=C _{n}^{r-1}a ^{n-r}b ^{r}$$
where n is the power, r is the rth term you need

didee · Answer

This is greek to me. How did you get to the 5th and 9th term?

callisto · Answer

Sorry i made mistakes there... it should be$$C _{r-1}^{n}a^{n-r}b^{r}$$
to find the 5th term, put n=8, r=5, a=1/x  and b= sqr root of x into that
to find the 9th term, put n=8, r=9, a=1/x  and b= sqr root of x into that

callisto · Answer

$$C _{r-1}^{n} = n! / [n-(r-1)]! (r-1)!$$

didee · Answer

still greek sorry. Lets assume I've never seen this before, which is true :-) where can I get some background on this. Any website suggestions maybe?

callisto · Answer

Perhaps this site: http://www.themathpage.com/aprecalc/binomial-theorem.htm I just learnt that from my teacher and my book , sorry :S

didee · Answer

Thanks so much, will check it out

callisto · Answer

Welcome :)