What have I done wrong here with this binomial expansion?

@topsquark there is the problem. After getting the values of j and k, I can’t substitute the values into the powers (1-j, 6-j) since I don’t have k’s as powers.
 
We want the coefficient of x^3 from [imath]\left (1 - \dfrac{x}{3} \right ) * (1 + 3x)^6[/imath]

Here is how I would proceed

I am interested in ONLY two terms of the expansion of [imath](1 + 3x)^6[/imath], namely the x squared and x cubed terms. WHY?

The x squared term is [imath]\dbinom{6}{2} * (3x)^2 * 1^4 = 15 * 9x^2 = 135x^2.[/imath]

The x cubed term is [imath]\dbinom{6}{3} * (3x)^3 * 1^3 = 20 * 27x^3 = 540x^3.[/imath]

Now what?
 
Kulla_9289 said:
@topsquark there is the problem. After getting the values of j and k, I can’t substitute the values into the powers (1-j, 6-j) since I don’t have k’s as powers.
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Huh? If you use j = 1 and k = 3, for example, then you have the term [imath]\binom{1}{1}(1)^1 \left ( -\dfrac{x}{3} \right ) ^0 \binom{6}{3}(1)^3 (3x)^3[/imath]. I don't understand where you are confused.

-Dan
 
topsquark said:
Huh? If you use j = 1 and k = 3, for example, then you have the term [imath]\binom{1}{1}(1)^1 \left ( -\dfrac{x}{3} \right ) ^0 \binom{6}{3}(1)^3 (3x)^3[/imath]. I don't understand where you are confused.

-Dan
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@tosquark What would I do with the other two numbers i.e. j=0 and k=4?

@Cubist Very much.
 
@Kulla_9289

You are making an easy problem very difficult for yourself by thinking about a double summation.

Did you read my post #42?

Can you give an answer (even if you expect it’s wrong) to the first question?
 
@topsquark Just tell me this; should I add the same values with different powers as k=0 and j=0?

@JeffM I can solve the question without double summation. This is just my curiosity.
 
Kulla_9289 said:
@topsquark Just tell me this; should I add the same values with different powers as k=0 and j=0?

@JeffM I can solve the question without double summation. This is just my curiosity.
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Any terms with j + k = 4 will be an [imath]x^3[/imath] term. So yes, just add the coefficients with possible j, k.

-Dan
 
@topsquark So basically, (summation notation..k=0)(summation notation..j=4) + (summation notation..k=1)(summation notation..j=3), am I right?
 
Kulla_9289 said:
@topsquark So basically, (summation notation..k=0)(summation notation..j=4) + (summation notation..k=1)(summation notation..j=3), am I right?
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Okay, that's where problem is!

[imath]\displaystyle \sum_{m = 0}^M a_m x^m \sum_{n = 0}^N b_n x^n = \sum_{m = 0}^M \sum_{n = 0}^N a_m b_n x^{m + n}[/imath]

[imath]= a_0 b_0 x^0 + (a_1 b_0 + a_0 b_1) x^1 + (a_2 b_0 + a_1 b_1 + a_0 b_2) x^2 + (a_3 b_0 + a_2 b_1 + a_1 b_2 + a_0 b_3) x^3 + \text{ ...}[/imath]

To get the coefficient of the power of x that you want you add the product of the coefficients of the factors you need. So to get the coefficient of the [imath]x^2[/imath] term you look at [imath]a_0 x^0[/imath] and [imath]b_1 x^2[/imath], [imath]a_1 x^1[/imath] and [imath]b_1 x^1[/imath], and [imath]a_2 x^2[/imath] and [imath]b_0 x^2[/imath]. Then you add the coefficients.

Here you have
topsquark said:
[math]\left ( 1 - \dfrac{x}{3} \right ) (1 + 3x )^6 = \sum_{j =0}^1 \binom{1}{j} (1)^j \left ( -\dfrac{x}{3} \right )^{1- j} \sum_{k = 0}^6 \binom{6}{k} (1)^k (3x)^{6 - k}[/math]
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and we need the j + k = 4 terms. That's j = 0, k = 4, and j = 1, k = 3. So add up those coefficients, ie. [imath]a_0 b_4 + a_1 b_3[/imath]. (You will not be summing from j = 0 to j = 1 or k = 0 to k = 4 or anything. Just plug in j = 0, k = 4 into the coefficients.)

-Dan
 
@topsquark I do not get 495x^3 as the answer. I got -585x^3 as the answer using the double summation. Would you please try it yourself?
 
Kulla_9289 said:
@topsquark I do not get 495x^3 as the answer. I got -585x^3 as the answer using the double summation. Would you please try it yourself?
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I might have been wrong. It happens. Could you please show your work so we can see if you did anything wrong? We can only help you if we know what you have done.

-Dan
 
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