***Some binomial questions*** (1 Viewer)

[norelle]

1.) Find r if 12C(3-r) = 12C(10-r) ANS:1

2.) Find the term independent of x ANS:-72
(x- 1/x)^10 (x+ 1/x)^4

3.) In the expansion of
(1+3x)^6 + (1+4x)^5 + (1-x)^10
Find the coefficients of x^8 ANS:45

4.) Given (1+px)^n = 1+20x+180x^2+kx^3+other terms involving higher powers of x, where n is a positive integer. Find n,p and k.
ANS: n=10 p=2 k=960

5.) In the expansion of (a+ x/b)^6, the coefficients of x and x^2 are 48 and 15 respectively. Find the possible values of a and b.
ANSL a=2, b=4 OR a=-2, b=-4


These are the questions I couldnt do
please help!! Thanks heapsss!!

 

[x.Exhaust.x]

Lol at the first question. To check, substitute r=1 into your calculator:



I don't think the answer is r=1.

 

[clintmyster]

provided you fix the mistake in the question, question 1 aint hard. Just expand using nCr = n!/r!(n-r)!

question 2: hmms haven't done this stuff in a while so i might have used the long method because i forgot about the easier one but anyways i expanded (x- 1/x)^10 and (x+ 1/x)^4 seperately using binomial theorem so that nC1 crap. Just be careful with the minus signs. Then i looked how to get x^0. This can be done with x^4/x^-4, x^2/x^-2,..., x^-4, x^4. Then i took the coefficients and just added them and got the answer.

question 3: similar logic, x^8 can only be found in (1-x)^10 because the other expansions have too low of a degree i.e. <8. Used the nCo...to the point you get the coefficient of x^8 to be +10C8 which is 45.

question 4: my its been a while since i last saw this question, like 6 whole months! ums basically expand using binomial LOL then for the second term which is (nC1)px it is equal to 20x. By a bit of manipulation i got p=20/n. Then for the third term nC2 p^2 x^2 = 180 x^2. sub that expression for p in, do some manipulation and you will get n = 10. Note I expand nC2 in this step so that an "n" would cancel out. So p from the first equation =20/10 = 2. Just do the same thing for k as i did above and you should get it.

question 5: (6C1)a^5(x/b) = 48, (6C2)a^4(x^2/b^2) = 15. square the first eqn and then divide the 2. the two answers come because its a^6 and you have to 6th root a number. I tried it really quickly and i didn't get a whole number but it does round to 2/-2 so that must be it. If someone could check this for me, that'd be great.

Hope that helps

 

[norelle]

omg for question 1, there was typing error,,

should be like this:
12C3r = 12C (10-r)

yes I did expand it,
but have no idea up to this step:

12! 12!
(12-3r)! (3r)! = (2+r)! (10-r)!

(12-3r)! (3r)! = (2+r)! (10-r)!

 

[MONKEYjulz]

I don't know if this is right. But i solved it using this method.
Using the law: nCk = nC(n-k)
Taking into consideration only the k bit, I went:
3r = 12 - (10 - r) which is k = n-k
3r = 2 + r
2r = 2
r = 1

Also works if you go:
12 - 3r = 10 - r

Hope you understand.

 

[MONKEYjulz]

Lol at the first question. To check, substitute r=1 into your calculator:



I don't think the answer is r=1.

LOL you stole my idea! =p

 

[MONKEYjulz]

question 2: hmms haven't done this stuff in a while so i might have used the long method because i forgot about the easier one but anyways i expanded (x- 1/x)^10 and (x+ 1/x)^4 seperately using binomial theorem so that nC1 crap. Just be careful with the minus signs. Then i looked how to get x^0. This can be done with x^4/x^-4, x^2/x^-2,..., x^-4, x^4. Then i took the coefficients and just added them and got the answer.

I did what you did and still couldn't get it.. the eff.. could u show the workin or something please

 

[clintmyster]

I did what you did and still couldn't get it.. the eff.. could u show the workin or something please

question 2: hmms haven't done this stuff in a while so i might have used the long method because i forgot about the easier one but anyways i expanded (x- 1/x)^10 and (x+ 1/x)^4 seperately using binomial theorem so that nC1 crap. Just be careful with the minus signs. Then i looked how to get x^0. This can be done with x^4/x^-4, x^2/x^-2,..., x^-4, x^4. Then i took the coefficients and just added them and got the answer.

(x+ 1/x)^4 = 4C0 x^4 + 4C1 x^2 + 4C2 x^0 + 4C3 x^-2 + 4C4 x^-4
(x- 1/x)^10 = 10C0 x^10 - 10C1 x^8 + 10C2 x^6 - 10C3 x^4 + 10C4 x^2 - 10C5 x^0 + 10C6 x^-2 - 10C7 x^-4 + 10C8 x^-6 - 10C9 x^-8 + 10C10 x^-10

so for x^4/x^-4, = 4C0 x -10C7 = -120
x^2/x^-2, = 4C1 x 10C6 = 840
x^0/x^-0 = 4C2 x -10C5 = -1512
x^-2/x^2 = 4C3 x 10C4 = 840
x^-4/x^4 = 4C4 x -10C3 = -120

add these up and you get -72

 

[norelle]

I don't know if this is right. But i solved it using this method.
Using the law: nCk = nC(n-k)
Taking into consideration only the k bit, I went:
3r = 12 - (10 - r) which is k = n-k
3r = 2 + r
2r = 2
r = 1

Also works if you go:
12 - 3r = 10 - r

Hope you understand.

--------
By following the nCk = nC(n-k),

I will get
12C(12-3r) = 12C(12-2+r)
12-3r = 2+r
10=4r
r = 5/2 ??!

 

[MONKEYjulz]

--------
By following the nCk = nC(n-k),

I will get
12C(12-3r) = 12C(12-2+r)
12-3r = 2+r
10=4r
r = 5/2 ??!

Why did u n-k both sides? you only do it to one side... expressed in the law itself.
Follow my steps written before, its clear.

Also, these two lines you wrote don't make sense. You must of typo'd
12C(12-3r) = 12C(12-2+r)
12-3r = 2+r

 

[MONKEYjulz]

question 2: hmms haven't done this stuff in a while so i might have used the long method because i forgot about the easier one but anyways i expanded (x- 1/x)^10 and (x+ 1/x)^4 seperately using binomial theorem so that nC1 crap. Just be careful with the minus signs. Then i looked how to get x^0. This can be done with x^4/x^-4, x^2/x^-2,..., x^-4, x^4. Then i took the coefficients and just added them and got the answer.

(x+ 1/x)^4 = 4C0 x^4 + 4C1 x^2 + 4C2 x^0 + 4C3 x^-2 + 4C4 x^-4
(x- 1/x)^10 = 10C0 x^10 - 10C1 x^8 + 10C2 x^6 - 10C3 x^4 + 10C4 x^2 - 10C5 x^0 + 10C6 x^-2 - 10C7 x^-4 + 10C8 x^-6 - 10C9 x^-8 + 10C10 x^-10

so for x^4/x^-4, = 4C0 x -10C7 = -120
x^2/x^-2, = 4C1 x 10C6 = 840
x^0/x^-0 = 4C2 x -10C5 = -1512
x^-2/x^2 = 4C3 x 10C4 = 840
x^-4/x^4 = 4C4 x -10C3 = -120

add these up and you get -72

Oops, i added x^0 and x^-0 coefficients, forgot to times em lol
Cheers man

 

[norelle]

I don't know if this is right. But i solved it using this method.
Using the law: nCk = nC(n-k)
Taking into consideration only the k bit, I went:
3r = 12 - (10 - r) which is k = n-k
3r = 2 + r
2r = 2
r = 1

Also works if you go:
12 - 3r = 10 - r

Hope you understand.

And is that a standard formula?
nCk = nC(n-k)
coz i cant find that in my textbk..

 

[norelle]

And is that a standard formula?
nCk = nC(n-k)
coz i cant find that in my textbk..

Oh ye I found it.

 

[MONKEYjulz]

And is that a standard formula?
nCk = nC(n-k)
coz i cant find that in my textbk..

YES! Of course.
Here's an example:
nCk = nC(n-k)
Let k = 2 and n = 12
12C2 = 12C(12-2)
12C2 = 12C10 which is true by calculator.
The way this theorem works is because in an expansion, the coefficients are mirrored both ways from the centre of the expansion.. you'll see in the pascal triangle it's like:
1 4 6 4 1 - power of 4
1 5 10 10 5 1 - power of 5

 

[clintmyster]

YES! Of course.
Here's an example:
nCk = nC(n-k)
Let k = 2 and n = 12
12C2 = 12C(12-2)
12C2 = 12C10 which is true by calculator.
The way this theorem works is because in an expansion, the coefficients are mirrored both ways from the centre of the expansion.. you'll see in the pascal triangle it's like:
1 4 6 4 1 - power of 4
1 5 10 10 5 1 - power of 5

Just keep in mind however that in most cases, there should be a way to get around this formula. Most people in my class would probably not even remember this existed. In some past papers i've seen, they make you prove the identity first so then you know you have to use it. But if worst comes to worst like in my case, just expand and cancel haha.

 

[Timothy.Siu]

well,
nCk=n!/k!(n-k)!

nC(n-k)=n!/(n-k)!k!

 

[norelle]

Thanks heaps everyone~~~!! I got all those questions now

but still stuck on this one:
Find the constant term in the expansion of (1+x^3 - 1/x^2 )^5

I did it in this way:
(1+x^3 - 1/x^2 )^5
= (1+ x^3 - x^-2)^5
= [ 1+ x^3(1- x^-5) ]^5

Tk = (5Ck) [(1)^5-k] [x^3 (1-x^-5)]^k
= (5Ck) x^3k (1- x^-5)^k

then what can I do up to this step ?

 

[Templar]

With this trinomial expression, you need something different

Note that the only way to get constant terms is with product of five 1's or two and three . There are 5C2 ways for the latter, and along with the minus sign, the coefficient is 1-10=9.