Totally lost on this one, here is the question
Which term in the expansion of $\displaystyle (1/2x^3-x^5)^8$is the constant?
$\displaystyle 1/2 is 1 over 2x^3$
The expansion looks like $\displaystyle \left( {\frac{1}{{2x^3 }} - x^5 } \right)^8 = \left( {2^{ - 1} x^{ - 3} - x^5 } \right)^8 = \sum\limits_{k=0}^{8} { \binom{8}{k} \left( {2^{ - 1} x^{ - 3} } \right)^k \left( { - x^5 } \right)^{8 - k} } $
So each individual term is $\displaystyle \binom{8}{k}\left( { - 1} \right)^{8 - k} \left( {2^{ - k} } \right)x^{40 - 8k} $.
What is the value of $\displaystyle k$ that give a constant term?
Hello, nuckers!
Which term in the expansion of .$\displaystyle \left(\frac{1}{2x^3}-x^5\right)^8$ is the constant?
The binomial expansion is:
. . $\displaystyle {8\choose8}\left(\frac{1}{2x^3}\right)^8 - {8\choose7}\left(\frac{1}{2x^3}\right)^7(x^5) + {8\choose6}\left(\frac{1}{2x^3}\right)^6(x^5)^2 - {8\choose5}\left(\frac{1}{2x^3}\right)^5\left(x^5\ right)^3 + \hdots $
. . . . . . $\displaystyle \hdots + {8\choose 2}\left(\frac{1}{2x^3}\right)^2\left(x^5)^6 - {8\choose1}\left(\frac{1}{2x^3}\right)(x^5)^7 + {8\choose0}(x^5)^8 $
The constant term is: .$\displaystyle -{8\choose5}\left(\frac{1}{2x^3}\right)^5\left(x^5\ right)^3 \;=\;-56\cdot\frac{1}{32x^{15}}\cdot x^{15} \;=\;-\frac{7}{4} $
You were told that the general term is $\displaystyle \binom{8}{k}\left( { - 1} \right)^{8 - k} \left( {2^{ - k} } \right)x^{40 - 8k} $.
The constant term corresponds to the term where x is raised to the power of zero since x^0 = 1. So you solve 40 - 8k = 0 and get k = 5. So the constant term is $\displaystyle \binom{8}{5}\left( { - 1} \right)^{8 - 5} \left( {2^{ - 5} } \right) = ....$
I think you meant the 6th term is the constant term free of x.
That corresponds to k=5 as the series starts at k=0.
Your original question seems to be only asking for the term number,
but it's safe to assume it's value is required, at least we learn more from evaluating it.