General term (1 Viewer)

[=)(=]

Can someone pls help with q 8a? I tried getting the general term of both elements and multiplying them but get the wrong answer.

 

[ExtremelyBoredUser]

Use the difference of squares for all the 4 terms. You should get (x+1/x)(x^2-1/x^2)^4

Then you would try to pinpoint where the x term would come:

1/x * x^2 = x
x * x^0 = x
Only two ways

Now you write the x^2 - 1/x^2 in general form, this is straightforward but I'll show anyways:

4CK * x^(8-4k) * (-1)^k

x^2 coefficient is;
8-4k = 2 therefore k = 3/2
k is an integer value hence no x^2 coefficient possible.

x^0 coefficient or the general term is;
8 - 4k = 0
k = 2

For x^0:
4C2 * (-1)^2 = 4C2 = 6

6 should be answer? Check with the textbook.

 

[=)(=]

Use the difference of squares for all the 4 terms. You should get (x+1/x)(x^2-1/x^2)^4

Then you would try to pinpoint where the x term would come:

1/x * x^2 = x
x * x^0 = x
Only two ways

Now you write the x^2 - 1/x^2 in general form, this is straightforward but I'll show anyways:

4CK * x^(8-4k) * (-1)^k

x^2 coefficient is;
8-4k = 2 therefore k = 3/2
k is an integer value hence no x^2 coefficient possible.

x^0 coefficient or the general term is;
8 - 4k = 0
k = 2

For x^0:
4C2 * (-1)^2 = 4C2 = 6

6 should be answer? Check with the textbook.

6 is the answer, I am a bit confused as to what you did in the first part with the difference of squares

 

[ExtremelyBoredUser]

6 is the answer, I am a bit confused as to what you did in the first part with the difference of squares

So this was the question;


You can unpack this;

(x+ 1/x) * (x+1/x)^4 * (x-1/x)^4

So now seeing it in this form, you can use difference of squares to simplify the equation.

Hence;
(x+1/x) * (x^2 - 1/x^2)^4

 

[=)(=]

ohh i see now thanks, would that be the same procedure for c?

 

[ExtremelyBoredUser]

ohh i see now thanks, would that be the same procedure for c?

Yep.

 

[=)(=]

thanks for the help really appreciate it

 

[=)(=]

Yep.

I tried part c but got (y+1/y)^3*(y^2-(1/y)^2)^7 what should I do from here?

 

[ExtremelyBoredUser]

I tried part c but got (y+1/y)^3*(y^2-(1/y)^2)^7 what should I do from here?

This one's pretty long I realised. You have to find the terms that will equal to -3. Get the Binomial Forms for both terms (y+ 1/y)^3 and (y^2 - 1/y^2)^7 and then find all combinations where they both equal to y^-3. If you're stuck from there, I'll give my solution.

Here are the forms to give you a headstart:

3CK y^(3-k) (1/y)^k
7CK * y^(14-2k) (-1)^k (1/y^2)^k

You can choose to substitute K with another variable if it gets confusing for example u, v or j when you are finding combinations.

 

[=)(=]

This one's pretty long I realised. You have to find the terms that will equal to -3. Get the Binomial Forms for both terms (y+ 1/y)^3 and (y^2 - 1/y^2)^7 and then find all combinations where they both equal to y^-3. If you're stuck from there, I'll give my solution.

Here are the forms to give you a headstart:

3CK y^(3-k) (1/y)^k
7CK * y^(14-2k) (-1)^k (1/y^2)^k

You can choose to substitute K with another variable if it gets confusing for example u, v or j when you are finding combinations.

Ohh okay so you basically got the general form for both elements and tried finding what values of k gives you y^-3 when you multiply them?

 

[ExtremelyBoredUser]

Ohh okay so you basically got the general form for both elements and tried finding what values of k gives you y^-3 when you multiply them?

Yep. For example y^-6 * y^3. For these type of questions, I usually draw up a table to find all the possible terms you can multiply to get y^-3. Then after that you just plug in the values to find k for each equation.

For example:

To get y^-6;
14-4k = -6
k = 5

Then you would try to find y^3 for the other equation

3-2k = 3
therefore k =0

7C5 * (-1)^5 multiplied by 3C0 is one combination

 

[=)(=]

Yep. For example y^-6 * y^3. For these type of questions, I usually draw up a table to find all the possible terms you can multiply to get y^-3. Then after that you just plug in the values to find k for each equation.

For example:

To get y^-6;
14-4k = -6
k = 5

Then you would try to find y^3 for the other equation

3-2k = 3
therefore k =0

7C5 * (-1)^5 multiplied by 3C0 is one combination

ohh okay got it, how would you make sure that you got all the terms which give y^-3?

 

[ExtremelyBoredUser]

ohh okay got it, how would you make sure that you got all the terms which give y^-3?

3CK y^(3-k) (1/y)^k
7CK * y^(14-4k) (-1)^k

Just see both of these expansions and then try to get a certain y value for each, remember they both multiply by each other.


Here are all the coefficients of the terms you are trying to find:

y^-1 * y^-2
y^1 * y^-4
y^3 * y^-6

You would just need to try to find the exponentials through plugging it into the exponents.
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For example:

To get y^-6;
14-4k = -6
k = 5

Then you would try to find y^3 for the other equation

3-2k = 3
therefore k =0

This is just multiplying the coefficients of y^-6 * y^3 as we are trying to find the coefficients for y^-3. You would do the same thing for the other 3. I'm not sure of any other method for these type of questions other than just doing them manually but this is the only one I've found that makes use of the binomial expression.