What have I done wrong here with this binomial expansion?

[Smith S.]

Hello,
I have been asked to find the $$x^3$$ term of the following expansion: $$(1-\frac{x}{3})(1+3x)^6$$. My answer to this was $$-585x^3$$. However, when I checked the answer scheme, it was $$495x^3$$. The error here is unknown to me; I have attached my steps to the answer. It would be appreciated if anyone could point out the error.

 

[pka]

I do not follow your work.
In the expansion of $\left(1+3x\right)^6$ do you have the term $540x^3~?$
$\left(1+3x\right)^6=729 x^6 + 1458 x^5 + 1215 x^4 + 540 x^3 + 135 x^2 + 18 x + 1$
You also need to multiply by $~1\&~\dfrac{-3}{x}$ to get any other $x^3$

 

[Smith S.]

Is there any other than to expand everything out? What would your approach to this problem be? I have sought out the general term of $$(1− 3 x )(1+3x) ^6$$ (in terms of binomial), then equated the x values to x^3.

 

[Dr.Peterson]

Hello,
I have been asked to find the $$x^3$$ term of the following expansion: $$(1-\frac{x}{3})(1+3x)^6$$. My answer to this was $$-585x^3$$. However, when I checked the answer scheme, it was $$495x^3$$. The error here is unknown to me; I have attached my steps to the answer. It would be appreciated if anyone could point out the error.

Please explain in words what you are doing there; parts of it look sensible, but I'm not sure of your notations. I expect to see you adding two numbers, corresponding to the two 3's you circled, but I don't see that.

 

[topsquark]

Hello,
I have been asked to find the $$x^3$$ term of the following expansion: $$(1-\frac{x}{3})(1+3x)^6$$. My answer to this was $$-585x^3$$. However, when I checked the answer scheme, it was $$495x^3$$. The error here is unknown to me; I have attached my steps to the answer. It would be appreciated if anyone could point out the error.

$$\left ( 1 - \dfrac{x}{3} \right ) (1 + 3x)^6 = \left ( 1 - \dfrac{x}{3} \right ) \sum_{j = 0}^6 \binom{6}{j} (1)^j (3x)^{6-j}$$
$$= \sum_{j = 0}^6 \binom{6}{j} (3x)^{6-j} - \sum_{j = 0}^6 \dfrac{x}{3} \binom{6}{j} (3x)^{6-j}$$
Now pick out the $$x^3$$ terms: That's j = 3 in the first sum and j= 4 in the second. The sum of these coefficients are
$\binom{6}{3} 3^3 - \binom{6}{4} \dfrac{1}{3} 3^2 = 495$

I can't follow your notation but apparently you didn't subtract.

-Dan

 

[Smith S.]

Please explain in words what you are doing there; parts of it look sensible, but I'm not sure of your notations. I expect to see you adding two numbers, corresponding to the two 3's you circled, but I don't see that.

What do you mean? When r=0, p=3 and r=1, p=2, x=3, since we are looking for the x^3 term. Then, I substituted those values into the general term of the expansion (Tr,p). How would you do it?

 

[Smith S.]

$$\left ( 1 - \dfrac{x}{3} \right ) (1 + 3x)^6 = \left ( 1 - \dfrac{x}{3} \right ) \sum_{j = 0}^6 \binom{6}{j} (1)^j (3x)^{6-j}$$
$$= \sum_{j = 0}^6 \binom{6}{j} (3x)^{6-j} - \sum_{j = 0}^6 \dfrac{x}{3} \binom{6}{j} (3x)^{6-j}$$
Now pick out the $$x^3$$ terms: That's j = 3 in the first sum and j= 4 in the second. The sum of these coefficients are
$\binom{6}{3} 3^3 - \binom{6}{4} \dfrac{1}{3} 3^2 = 495$

I can't follow your notation but apparently you didn't subtract.

-Dan

How did you get j=3, and j=4?

 

[JeffM]

Suppose we want to know the coefficient of x cubed in the expansion of $(3 - 4x) (2 + x)^4$.

The first thing I would do is

$$(3 - 4x) (2 + 3x)^4 = 3(2 + x)^4 - 4x(3x + 2)^4$$.

The next thing I would do is to realize that I need the x cubed term from the binomial expansion to multiply by 3 but the x squared term to multiply by - 4x.

$$(3x + 2)^4 = \text {irrelevant} + \dbinom{4}{2} * (3x)^2 * 2^2 + \binom{4}{3} * (3x)^1 * 2^3 + \text {irrelevant}.$$
So my two terms are 3 * 6 * 9 * 4 = 648 and
- 4 * 4 * 3 * 8 = - 384.

The coeffecient of x^3 is 264.

 

[Smith S.]

Suppose we want to know the coefficient of x cubed in the expansion of $(3 - 4x) (2 + x)^4$.

The first thing I would do is

$$(3 - 4x) (2 + 3x)^4 = 3(2 + x)^4 - 4x(3x + 2)^4$$.

The next thing I would do is to realize that I need the x cubed term from the binomial expansion to multiply by 3 but the x squared term to multiply by - 4x.

$$(3x + 2)^4 = \text {irrelevant} + \dbinom{4}{2} * (3x)^2 * 2^2 + \binom{4}{3} * (3x)^1 * 2^3 + \text {irrelevant}.$$
So my two terms are 3 * 6 * 9 * 4 = 648 and
- 4 * 4 * 3 * 8 = - 384.

The coeffecient of x^3 is 264.

Why did you factor out 3 and -4?

 

[Deleted member 4993]

Why did you factor out 3 and -4?

Please work with pencil and paper (even chalk and slate) will do.

3 and (-4) multiplicative constants coming from (3 - 4x). Those are constants - would not affect the "locations" of the required coefficients

 

[BigBeachBanana]

How did you get j=3, and j=4?

It's basic algebra. Ask yourself what value of $j$ will give you 3?

For the first summation, $6-j=3 \implies j=3\\$
For the second summation, $1 + (6-j)=3 \implies j=4$

 

[Smith S.]

@topsquark What if there was a power like $$(1− 3 x )^5(1+3x) ^6$$?

 

[Steven G]

How do you get an x^3 term from (1-x/3)(1+3x)^6?

(1-x/3)(1+3x)^6 = 1(1+3x)^6 - (x/3)(1+3x)^6

From the factor 1(1+3x)^6, you will need the x^3 term from (1+3x)^6
From the factor (x/3)(1+3x)^6, you will need the x^2 term from (1+3x)^6

 

[topsquark]

@topsquark What if there was a power like $$(1− 3 x )^5(1+3x) ^6$$?

Then you would have to use a double summation, which would get a tad ugly. But it's doable:
$$(1 - 3x)^5 (1 + 3x)^6 = \sum_{j = 0}^5 \binom{5}{j} (1)^j (-3x)^{5 - j} \sum_{k = 0}^6 \binom{6}{k} (1)^k (3x)^{6 - k}$$
$$= \sum_{j = 0}^5 \sum_{k = 0}^6 \binom{5}{j} \binom{6}{k} (-3x)^{5 - j} (3x)^{6 - k}$$
$$= \sum_{j = 0}^5 \sum_{k = 0}^6 \binom{5}{j} \binom{6}{k} (-1)^{5 - j} (3)^{11 - j - k} x^{11 - j - k}$$
To get the coefficient of $x^3$ we would need to add all of the coefficients with j and k values such that 3 = 11 - j - k.

-Dan

 

[Smith S.]

Then I could create a two-way table to find the values of j and k that add up to 3; something like the attachment above. I attempted to use this way (double summation one you just showed) to solve the $$(1− 3 x )(1+3x)^6$$, but it turns out, it didn't work for power of 1. Am I right?

 

[lookagain]

$$(3 - 4x) (2 + 3x)^4 = 3(2 + x)^4 - 4x(3x + 2)^4$$.

Did you intend $\displaystyle \ (3 - 4x)(2 + 3x)^4 \ = \ 3(2 + 3x)^4 \ - \ 4x(3x + 2)^4 \ ?$

 

[pka]

@topsquark What if there was a power like $$(1− 3 x )^5(1+3x) ^6$$?

$(1-3x)^5(1+3x)^6\\\left((1-3x)^5(1+3x)^5\right)(1+3x)\\\left((1-9x^2)^5\right)(1+3x)\\= -177147 x^{11} - 59049 x^{10} + 98415 x^9 + 32805 x^8 - 21870 x^7 - 7290 x^6 + 2430 x^5 + 810 x^4 - 135 x^3 - 45 x^2 + 3 x + 1$ SEE HERE

$$$$$$$$

 

[topsquark]

Then I could create a two-way table to find the values of j and k that add up to 3; something like the attachment above. I attempted to use this way (double summation one you just showed) to solve the $$(1− 3 x )(1+3x)^6$$, but it turns out, it didn't work for power of 1. Am I right?

Using a double sum for this is like using a nuke to crack a walnut but what the heck.
$$(1 - 3x)(1 + 3x)^6 = \sum_{j = 0}^1 \binom{1}{j} (1)^j (-3x)^{1-j} \sum_{k = 0}^6 \binom{6}{k} (1)^k (3x)^{6 - k}$$
$$= \sum_{j = 0}^1 \sum_{k = 0}^6 \binom{1}{j} \binom{6}{k} (-3x)^{1-j} (3x)^{6 - k}$$
$$= \sum_{j = 0}^1 \sum_{k = 0}^6 \binom{1}{j} \binom{6}{k} (-1)^{1 - j} 3^{7 -j - k} x^{7 - j - k}$$
For the $x^3$ term, pick out j, k such that 7 - j - k = 3. Well, j only goes from 0 to 1, so when j = 0, k = 4 and when j= 1 then k = 3. So the coefficient of $x^3$ will be
$\binom{1}{0} \binom{6}{4} (-1)^{1 - 0} 3^{7 -0 - 4} + \binom{1}{1} \binom{6}{3} (-1)^{1 - 1} 3^{7 -1 - 3}$

But it's simpler just to distribute:
$$(1 - 3x)(1 + 3x)^6 = \sum_{k = 0}^6 \binom{6}{k} (1)^k (3x)^{6 - k} + \sum_{k = 0}^6 \binom{6}{k} (1)^k (-3x) (3x)^{6 - k}$$
and go from there.

-Dan

 

[Deleted member 4993]

Using a double sum for this is like using a nuke to crack a walnut but what the heck

Or using Quantum Physics to kill a cat......

In Bengal, there is a proverb - don't fire a cannon to kill a mosquito on the wall........(you'll hurt yourself)....

 

[lookagain]

In Bengal, there is a proverb - don't fire a cannon to kill a mosquito on the wall........(you'll hurt yourself)....

You'll more than kill the mosquito. You'll put a hole through the wall where the
mosquito was.