help with finding total coefficient for x^0 for binomial expansion of (X^2+1/x)^12.. i get the coefficient is 12C0 x^2^0 ((((or 1)))) * 1/x^12... but my book says 495...how do i get that?

Question

campbell_st · Answer

its when 

$$(x^2)^{12-r}(x^{-1})^r = x^0$$
equating the powers

24 - 2r - r = 0

24 = 3r

r = 8

so then you need to find 
$$^{12}C _{8} (x^2)^{12 - 8} (x^{-1})^8$$

gives $$495(x^8)(x^{-8}) = 495$$

anonymous · Answer

thank you, but why if its the coefficient of x^0 term, why is it not just the final term?

anonymous · Answer

or does it mean when simply all X's will cancel out from the binomial

campbell_st · Answer

the final term is 

$$^{12}C _{12}(x^2)^0(x^{-1})^12 = ^{12}C _{12}x^{-12}$$

campbell_st · Answer

its when the product of the terms results in a power of zero... 

remember multiplying the same pronumeral.... add the powers...

anonymous · Answer

so if i wanted when x^4, just set it equal to 4. awesome thanks so much

campbell_st · Answer

yep it would be 24 - 2r - r = 4

or 24 - 3r = 4 but it you can't get 4

the powers will be 24, 21, 18, 15, ... to -12

etc

anonymous · Answer

This is the whole expansion
$$ \left(x^2+\frac{1}{x}\right)^{12}=
x^{24}+12 x^{21}+66 x^{18}+220 x^{15}+\495 x^{12}+\frac{1}{x^{12}}+\792
   x^9+\frac{12}{x^9}+924 x^6+\frac{66}{x^6}+792 x^3+\\frac{220}{x^3}+495
$$

anonymous · Answer

The method described by @campbell_st is the way to do it.