anonymous
  • anonymous
(9+1)^(1/2) Find the first 4 terms of this using binomial expansion,
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sqrt{10}\]?
anonymous
  • anonymous
It's like (1/2)C0(a)^(1/2)(b)^(0) + (1/2)C1(a)^(-1/2)(b)^(1) and so on for 4 terms. I just don't know how to combine it.
anonymous
  • anonymous
oooh i see, maybe like \[(1+9)^{\frac{1}{2}}=1+\frac{1}{2}\times 9+\frac{\frac{1}{2}\times -\frac{1}{2}}{2!}9^2+..\]

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amistre64
  • amistre64
\[(1+a)^n=1+na+\frac{n(n-1)}{2!}a^2+\frac{n(n-1)(n-2)}{3!}a^3\] if i remember it correctly
anonymous
  • anonymous
yes that is it i am sure (more or less) but because you are dealing with fractional exponent it goes on forever. and needs to look like \[(1+x)^n\]
anonymous
  • anonymous
ummm. wut?
amistre64
  • amistre64
your going to have to be more specific than that ....
anonymous
  • anonymous
how did you get that string of numbers?
amistre64
  • amistre64
its the definition of a binomial expansion for rational exponents
amistre64
  • amistre64
im not quite sure how they come up with it yet, but i recall seeing it as such
anonymous
  • anonymous
that is not going to work for fractional exponents
anonymous
  • anonymous
read here for a brief explanation
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amistre64
  • amistre64
\[\binom{n}{1}=\frac{n}{1!}\] \[\binom{n}{2}=\frac{n(n-1)}{2!}\] \[\binom{n}{3}=\frac{n(n-1)(n-2)}{3!}\] etc...
amistre64
  • amistre64
those are textbook definitions i beleive
anonymous
  • anonymous
could you expand of (9+1)^(1/2) to 4 terms? I think that'd make more sense to me than just variables O_Ollll
amistre64
  • amistre64
satellite posted it above
amistre64
  • amistre64
4th post down

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