Petrus said:Decide constant term in https://webwork.math.su.se/webwork2_files/tmp/equations/67/172c674f7e20bbdefabddc24741bac1.png.
I have problem with this one, I can't find any exemple about this one in my book, any advice would be great:)
Thanks!chisigma said:If You apply the bynomial sum...
$\displaystyle (a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ a^{k}\ b^{n-k}$ (1)
... with $a=3\ x^{3}$, $a=- \frac{4}{x}$ and $n=20$ the constant term is for $\displaystyle 3\ k = 20 - k \implies k=5$ so that the connstant term is... $\displaystyle c= \binom{20}{5}\ 3^{5}\ (-4)^{15}$ (2)Kind regards
$\chi$ $\sigma$
Hello,chisigma said:If You apply the bynomial sum...
$\displaystyle (a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ a^{k}\ b^{n-k}$ (1)
... with $a=3\ x^{3}$, $a=- \frac{4}{x}$ and $n=20$ the constant term is for $\displaystyle 3\ k = 20 - k \implies k=5$ so that the connstant term is... $\displaystyle c= \binom{20}{5}\ 3^{5}\ (-4)^{15}$ (2)Kind regards $\chi$ $\sigma$
MarkFL said:The general term, as given by the binomial theorem, is:
\(\displaystyle {20 \choose k}\left(3x^3 \right)^{20-k}\left(-\frac{4}{x} \right)^k=3^{20-k}(-4)^k{20 \choose k}x^{3(20-k)}x^{-k}=3^{20-k}(-4)^k{20 \choose k}x^{4(15-k)}\)
What must the exponent on $x$ be in order for the term to be a constant?
Replying to this for some latex mall:)MarkFL said:The general term, as given by the binomial theorem, is:
\(\displaystyle {20 \choose k}\left(3x^3 \right)^{20-k}\left(-\frac{4}{x} \right)^k=3^{20-k}(-4)^k{20 \choose k}x^{3(20-k)}x^{-k}=3^{20-k}(-4)^k{20 \choose k}x^{4(15-k)}\)
What must the exponent on $x$ be in order for the term to be a constant?
That last one i did not know, why would i like to write it on that form? There is so many diffrent form to write but in exam how would you answer?MarkFL said:That's one way to express the constant term. It is a very large number, so I think I would leave it as:
\(\displaystyle C=-3^5\cdot4^{15}{20 \choose 15}\)
or even
\(\displaystyle C=-\left(3\cdot4^3 \right)^5{20 \choose 5}\)
Do you know the identity \(\displaystyle {n \choose r}={n \choose n-r}\)?
Hello Mark,MarkFL said:You are right that there are many forms to write this constant, and unless you are directed to write it a certain way, I suppose it is up to you how you choose to express the result.
I would choose the last form I gave simply because it's just ever so slightly more compact than the first form I gave.
In the form you gave, you would certainly want to simplify $(20-15)!$ to $5!$.
As a follow-up, can you demonstrate that the combinatorial identity I cited is true?
Thanks Mark!MarkFL said:Petrus has asked that I give him another similar problem for practice that he can try to solve. I decided to post it here for the benefit of others perhaps reading this topic who are wanting help with this kind of problem:
Find the constant term in the expansion of:
\(\displaystyle x^3\left(\frac{x^3}{y^2}-\frac{3y}{x^2} \right)^9\)
A constant term in a binomial expansion is a term that does not contain any variables. It is a number that remains the same throughout the expansion.
To find the constant term in a binomial expansion, you can use the formula (a + b)^n, where a and b are the terms in the binomial and n is the exponent. The constant term will be the term that does not contain any variables, so it will be the last term in the expansion.
The constant term is important because it represents the value of the binomial when both of its variables are equal to zero. This can be useful in solving certain equations or in finding the y-intercept of a graph.
The constant term is the only term in the binomial expansion that does not contain any variables. It is also the smallest term in the expansion, as it has the lowest exponent. The other terms in the expansion will have increasing exponents and may contain variables.
Yes, there are two special cases for the constant term in a binomial expansion. If the exponent n is even, the constant term will be the middle term in the expansion. If the exponent is odd, there will be no constant term in the expansion.