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?

$\displaystyle 1/2 is 1 over 2x^3$

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- Jul 25th 2010, 02:05 PMnuckersfinding constant term of expansion using binomial theorem
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?

$\displaystyle 1/2 is 1 over 2x^3$ - Jul 25th 2010, 02:17 PMpickslides
In the expansion $\displaystyle (x+y)^n$ the $\displaystyle (r+1)\text{th}$ term is given by $\displaystyle ^nC_rx^{n-r}y^r$ your job is to apply this for $\displaystyle x^0$

- Jul 25th 2010, 02:41 PMnuckers
- Jul 25th 2010, 02:47 PMPlato
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? - Jul 25th 2010, 02:59 PMnuckers
- Jul 25th 2010, 03:17 PMPlato
- Jul 25th 2010, 04:23 PMnuckers
- Jul 25th 2010, 05:44 PMmr fantastic
- Jul 25th 2010, 06:37 PMnuckers
- Jul 25th 2010, 08:18 PMSoroban
Hello, nuckers!

Quote:

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} $

- Jul 25th 2010, 08:29 PMnuckers
- Jul 26th 2010, 02:48 AMmr fantastic
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 $\displaystyle \binom{8}{5}\left( { - 1} \right)^{8 - 5} \left( {2^{ - 5} } \right) = ....$ - Jul 26th 2010, 03:33 AMArchie Meade
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.