# finding constant term of expansion using binomial theorem

• Jul 25th 2010, 02:05 PM
nuckers
finding constant term of expansion using binomial theorem
Totally lost on this one, here is the question

Which term in the expansion of $(1/2x^3-x^5)^8$is the constant?

$1/2 is 1 over 2x^3$
• Jul 25th 2010, 02:17 PM
pickslides
In the expansion $(x+y)^n$ the $(r+1)\text{th}$ term is given by $^nC_rx^{n-r}y^r$ your job is to apply this for $x^0$
• Jul 25th 2010, 02:41 PM
nuckers
Quote:

Originally Posted by pickslides
In the expansion $(x+y)^n$ the $(r+1)\text{th}$ term is given by $^nC_rx^{n-r}y^r$ your job is to apply this for $x^0$

i understand the formula, but i'm having a tough time trying to break it down, do i not have to get the bases the same to make it easier, thats the part i'm struggling with
• Jul 25th 2010, 02:47 PM
Plato
The expansion looks like $\left( {\frac{1}{{2x^3 }} - x^5 } \right)^8 = \left( {2^{ - 1} x^{ - 3} - x^5 } \right)^8 = \sum\limits_{k=0}^{8} { \binom{8}{k} \left( {2^{ - 1} x^{ - 3} } \right)^k \left( { - x^5 } \right)^{8 - k} }$

So each individual term is $\binom{8}{k}\left( { - 1} \right)^{8 - k} \left( {2^{ - k} } \right)x^{40 - 8k}$.

What is the value of $k$ that give a constant term?
• Jul 25th 2010, 02:59 PM
nuckers
Quote:

Originally Posted by Plato
The expansion looks like $\left( {\frac{1}{{2x^3 }} - x^5 } \right)^8 = \left( {2^{ - 1} x^{ - 3} - x^5 } \right)^8 = \sum\limits_{k=0}^{8} { \binom{8}{k} \left( {2^{ - 1} x^{ - 3} } \right)^k \left( { - x^5 } \right)^{8 - k} }$

So each individual term is $\binom{8}{k}\left( { - 1} \right)^{8 - k} \left( {2^{ - k} } \right)x^{40 - 8k}$.

What is the value of $k$ that give a constant term?

if i'm reading this right, then i will have to say 6 is the constant term
• Jul 25th 2010, 03:17 PM
Plato
Quote:

Originally Posted by nuckers
if i'm reading this right, then i will have to say 6 is the constant term

No indeed. What value of $k$ gives $x^0$?
Use that value to answer the question.
• Jul 25th 2010, 04:23 PM
nuckers
Quote:

Originally Posted by Plato
No indeed. What value of $k$ gives $x^0$?
Use that value to answer the question.

Thanks for the help, but for some reason i just can't seem to understand it,
• Jul 25th 2010, 05:44 PM
mr fantastic
Quote:

Originally Posted by nuckers
Thanks for the help, but for some reason i just can't seem to understand it,

Go back and read post # 4. Do you understand that x^0 is equal to 1? Therefore the constant term corresponds to the term where the power of x is equal to 0.
• Jul 25th 2010, 06:37 PM
nuckers
Quote:

Originally Posted by mr fantastic
Go back and read post # 4. Do you understand that x^0 is equal to 1? Therefore the constant term corresponds to the term where the power of x is equal to 0.

i do understand that x^0, but in order to get that i need to make the coefficient equal 1, but thats what i'm having problems with, how do i get rid of the 2?
• Jul 25th 2010, 08:18 PM
Soroban
Hello, nuckers!

Quote:

Which term in the expansion of . $\left(\frac{1}{2x^3}-x^5\right)^8$ is the constant?

The binomial expansion is:

. . ${8\choose8}\left(\frac{1}{2x^3}\right)^8 - {8\choose7}\left(\frac{1}{2x^3}\right)^7(x^5) + {8\choose6}\left(\frac{1}{2x^3}\right)^6(x^5)^2 - {8\choose5}\left(\frac{1}{2x^3}\right)^5\left(x^5\ right)^3 + \hdots$

. . . . . . $\hdots + {8\choose 2}\left(\frac{1}{2x^3}\right)^2\left(x^5)^6 - {8\choose1}\left(\frac{1}{2x^3}\right)(x^5)^7 + {8\choose0}(x^5)^8$

The constant term is: . $-{8\choose5}\left(\frac{1}{2x^3}\right)^5\left(x^5\ right)^3 \;=\;-56\cdot\frac{1}{32x^{15}}\cdot x^{15} \;=\;-\frac{7}{4}$

• Jul 25th 2010, 08:29 PM
nuckers
Quote:

Originally Posted by Soroban
Hello, nuckers!

The binomial expansion is:

. . ${8\choose8}\left(\frac{1}{2x^3}\right)^8 - {8\choose7}\left(\frac{1}{2x^3}\right)^7(x^5) + {8\choose6}\left(\frac{1}{2x^3}\right)^6(x^5)^2 - {8\choose5}\left(\frac{1}{2x^3}\right)^5\left(x^5\ right)^3 + \hdots$

. . . . . . $\hdots + {8\choose 2}\left(\frac{1}{2x^3}\right)^2\left(x^5)^6 - {8\choose1}\left(\frac{1}{2x^3}\right)(x^5)^7 + {8\choose0}(x^5)^8$

The constant term is: . $-{8\choose5}\left(\frac{1}{2x^3}\right)^5\left(x^5\ right)^3 \;=\;-56\cdot\frac{1}{32x^{15}}\cdot x^{15} \;=\;-\frac{7}{4}$

Thanks so much, that puts it more in perspective, i wasn't even doing it that way, i was trying to make the bases the same, as they showed me, never even thought of doing it that way, thanks so much
• Jul 26th 2010, 02:48 AM
mr fantastic
Quote:

Originally Posted by nuckers
Thanks so much, that puts it more in perspective, i wasn't even doing it that way, i was trying to make the bases the same, as they showed me, never even thought of doing it that way, thanks so much

You were told that the general term is $\binom{8}{k}\left( { - 1} \right)^{8 - k} \left( {2^{ - k} } \right)x^{40 - 8k}$.

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 $\binom{8}{5}\left( { - 1} \right)^{8 - 5} \left( {2^{ - 5} } \right) = ....$
• Jul 26th 2010, 03:33 AM