PLEASE HELP REALLY NEED HELP!!!!!!!!!!!!!!!!!!!!!!!!! WILL GIVE MEDAL!!!!!!!!!!!!!!!!!!!!!!!!!!
Find the constant term of expansion of the binomial (6x+1/3x)^6?
@amistre64 please help
might need some clarification on the notation, and a definition on what a constant term of expansion refers to
\[\frac1{3x}\] or \[\frac13x\]
So what to do after that?
Generally if you have (x^p + x^q)^n then you make things easier by taking one term outside the bracket: x^(np)[1 + x(q-p)]^n = x^(np)[1 + C(n,1)x^(q-p) + C(n,2)x^(2(q-p)) + C(n,3)x^(3(q-p)) +.. The constant term would be C(n, r) where np + r(q - p) = 0 if an integer value of r satisfies this equation. http://mathforum.org/library/drmath/view/56428.html
amisster we can't find the answer if it is \[{1x/3}\]
the answer is 4th term
I am sorry I don't understand at all. Can you please explain this to me in depth. This is extra credit so it is essential that I get it right or my grade in math will be a B-!
Any help is appreciated.
then (r+1) term of the expantion (a+b)^n is c(n,r)a^(n-r)*b^r
\[(6x+\frac13x^{-1})^6\] \[(6x)^{6}(1+\frac1{18}x^{-2})^6\]
@amistre64 I will give the medal to @tony115 because you can't get much more higher than 99, is that okay with you?
fine by me, as long as they can elucidate the solution :)
what is the solution?
please say that do you understand my recent comment
if we take and actually expand it out we get this: 46656 x^6 +1/(729 x^6) +15552 x^4 +4/(27 x^4) +2160 x^2 +20/(3 x^2) +160 im wondering of the 160 is what the question is asking for
I am sorry i didnt understand that last post @tony115
yes that the value of 4th term
in the expantion of \[(a+b)^n\] (r+1)th term is obtained by c(n,r)\[a^{n-r}*b^{r}\]
sorry,\[c(n,r)a^{n-r}*b^{r}\]
so its the x^0 term ... i was confused on what it was actually asking for :)
160 was the right answer, thanks guys or gals!!!!!!
rob, you interpreted the question a little off. a = x^-1
bye
one more question
Expand (2x - 1)^5.
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