Thread: finding constant term of expansion using binomial theorem

1. finding 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?

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

2. 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 $\displaystyle x^0$

3. Originally Posted by pickslides
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 $\displaystyle x^0$
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

4. 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?

5. Originally Posted by Plato
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?

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

6. 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 $\displaystyle k$ gives $\displaystyle x^0$?
Use that value to answer the question.

7. Originally Posted by Plato
No indeed. What value of $\displaystyle k$ gives $\displaystyle x^0$?
Use that value to answer the question.
Thanks for the help, but for some reason i just can't seem to understand it,

8. 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.

9. 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?

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

11. Originally Posted by Soroban
Hello, nuckers!

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

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

12. 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 $\displaystyle \binom{8}{k}\left( { - 1} \right)^{8 - k} \left( {2^{ - k} } \right)x^{40 - 8k}$.

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 $\displaystyle \binom{8}{5}\left( { - 1} \right)^{8 - 5} \left( {2^{ - 5} } \right) = ....$

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

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