Didee
  • Didee
The middle term in the expansion of (1/x + sqr root of x)^8 is?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
do you mean this?? \[\Large \left(\frac1x+\sqrt x\right)^8\]
Callisto
  • Callisto
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
  • anonymous
\[\dbinom{8}{4}(\frac{1}{x})^4\sqrt{x}^4\] is a start

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Didee
  • Didee
@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
  • Callisto
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
  • Didee
This is greek to me. How did you get to the 5th and 9th term?
Callisto
  • Callisto
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
  • Callisto
\[C _{r-1}^{n} = n! / [n-(r-1)]! (r-1)!\]
Didee
  • Didee
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
  • Callisto
Perhaps this site: http://www.themathpage.com/aprecalc/binomial-theorem.htm I just learnt that from my teacher and my book , sorry :S
Didee
  • Didee
Thanks so much, will check it out
Callisto
  • Callisto
Welcome :)

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