Question

Can someone explain how to find the constant term using the binomial theorem?

Answer 1

\[(6x + \frac{ 1 }{ 3x })^6\]

Answer 2

\[(a+b)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k\]
Here, \(a=6x\), \(b=\dfrac{1}{3x}\), and \(n=6\).

The constant term will occur when the powers of \((6x)^{6-k}\) and \(\left(\dfrac{1}{3x}\right)^k=(3x)^{-k}\) cancel out. This happens when \(6-k+(-k)=0~\Rightarrow~k=3\).

So, you'd plug in \(k=3\) into the summation formula:
\[\begin{pmatrix}6\\3\end{pmatrix}(6x)^{6-3}\left(\frac{1}{3x}\right)^3=\cdots\]

Answer 3

Uhh, I don't understand what this:

symbol means

Answer 4

@SithsAndGiggles

Answer 5

It's a summation.

For example,
\[\sum_{n=0}^2n=0+1+2\]
Have you been formally introduced to the binomial theorem?

Answer 6

Yes, with combinations, where (x+y)^n = nC0x^n + nC1x^(n-1)y, +...... nCny^n

Answer 7

Ah, well my summation notation is a way of saying the same thing.

\[\sum_{k=0}^n {}_nC_{k}x^{n-k}y^k={}_nC_0x^n+{}_nC_1x^{n-1}y+\cdots+{}_nC_ny^n\]
\(k\) is called an index, where \(k=0,1,2,3,...\)

Answer 8

Do you understand how to find the constant term though?

Answer 9

Sort of

Answer 10

Let me try to do another example

Answer 11

\[(x^2 + \frac{ 1 }{ x })^4\]

Answer 12

a = x^2 b = 1/x n = 4

Answer 13

and k would be found by doing 4-k + -k = 0 = -2k = 4 = k = -2?

Answer 14

\(k\) must be positive!
\[4-k+(-k)=0~\Rightarrow~4=2k~\Rightarrow~k=2\]

Answer 15

Whoops, my mistake, I forgot to flip the sign when I moved it over.

Answer 16

Also, I got 160 for the first example? Is that correct?

Answer 17

So, if \(k=2\), then the constant term will be
\[{}_4C_2\left(x^2\right)^{4-2}\left(\frac{1}{x}\right)^2=\cdots\]
hmm, looks like you made a mistake here. This gives you a nonconstant term. Are you supposed to find a constant term?

And yes, you're right about 160.

Answer 18

Here is a screenshot of the question
http://gyazo.com/0ae4153301d80ff9a17baf1d5c71ca5e

Answer 19

Okay, the \(x^2\) must be tripping you up.

From the formula, you have
\[\left(x^2+\frac{1}{x}\right)^4={}_4C_0(x^2)^4+{}_4C_1(x^2)^3\left(\frac{1}{x}\right)+{}_4C_2(x^2)^2\left(\frac{1}{x}\right)^2\\
~~~~~~~~~~~~~~~~~~~~~~~+{}_4C_3x^2\left(\frac{1}{x}\right)^3+{}_4C_4\left(\frac{1}{x}\right)^4\]
Each term will contain a power of \(x\), so there is not constant term. Well, there is, but it's 0.

Answer 20

Each term will be of the form
\[{}_4C_k\left(x^2\right)^{4-k}\left(\frac{1}{x}\right)^k\]
Rewriting a bit, you have
\[{}_4C_k\left(x^2\right)^{4-k}\left(x^{-1}\right)^k\\
{}_4C_k\left(x\right)^{2(4-k)}\left(x\right)^{-k}\\
{}_4C_k\left(x\right)^{2(4-k)-k}\]
The constant term would appear when the exponent is 0:
\[2(4-k)-k=0\\
8-3k=0\\
8=3k\]
Although there is a solution for \(k\) here, keep in mind that \(k\) must be a whole number (\(k=0,1,2,3,...\)). So, the constant term is zero.

Answer 21

Okay, I want to try one more problem to make sure I understand.

Answer 22

\[\left( x^2 + \frac{ 1 }{ x } \right)^6\]

Answer 23

For this one, you can use the same reasoning as the previous problem. The only difference is that \(n=6\), and not 4.

Answer 24

6Ck (x^2)^(6-k) * (1/x)^k

Answer 25

2(6-k) - k = 12-2k-k
12 = 3k
k = 4

Answer 26

Yes, so now you just plug in \(k=4\) into \(
{}_6C_k\left(x\right)^{2(6-k)-k}\)

Answer 27

I got 15, is that correct?

Answer 28

Yes

Answer 29

Thank a ton, there are quite a bit of questions like this, and you've help me a lot

Answer 30

You're welcome!