Binomial Theorm T__T (1 Viewer)

[syriangabsta]

hmm, ive been given a revision sheet by my teacher, and i think im stuck on a few questions. Could anyone help me? kthx

okay the first question..is..

Find the coefficient of x^2 in the expansion of ( x + 1/x)^3 (x - 1/x)^5..

All i know is that you have to expand it..and yeh, please help.

The next question is...

In the expansion of (3x - 4y)^12:

What is the coefficient of the term in wihch the powers of x and y are equal..

i found out that when powers of x and y are equal, its the 7th term, so...T (6+1). and got 12C6 . (3x)^6 . (-4y)^6..

after expanding and similpyfying, i removed the x and y, and got...

12C6 . (-4)^6 . 3^6...this got me an answer of 2.579049216 E09
could anyone tell me if this is right..i thought it was wrong coz the number is so big lol. If my answer is wrong, could someone show me how they did it too?kthx

umm another question for now..

In the expansion of (3 + 4x )^n the coefficients of x^2 and x^3 and in the ratio 3:4. Find the value of n.

I got an answer for this question but have no clue if its correct.

I first expanded part of it..untill i got to x^2 and x^3...which were..

nC2 . 3^n-2 . 16x^2 (i) and nC3 . 3^n-3 . 64x^3 (ii)

I put that into the form (i)/(ii) and made it equal to 3/4..

After simplifying, i got..

3n-9 . 3^-5 . 1/4 = 3/4

and use simple algebra, to find n to be 246. Is this right? i couldnt sub into calculator T__T numbers are too big. help plz.

I think ill have more questions later

 

[aalex]

Q1

Find the coefficient of x^2 in the expansion of ( x + 1/x)^3 (x - 1/x)^5.

First we write (x - 1/x)^5 = (x - 1/x)^3. (x - 1/x)^2 and then substitute it in the relation above and use the property a^n.b^n = (ab)^n and (a + b).(a - b) = a^2 – b^2<O</O

we can rewrite:
<O
(x^2 - 1/x^2)3.(x - 1/x)^2 and now we expand the 2 brackets(cube and square):
<O
(x^6 - 3x^4.1/x^2 + 3x^2.1/x^4-1/x^6).(x^2 – 2 – 1/x^2) = <O
(x^6 - 3x^2 + 3.1/x^2-1/x^6).(x^2 – 2 – 1/x^2)
<O</O

The only term that has x to the power of 2 is obtained from (-3x^2).(-2) = 6x^2, so the coefficient is 6.


Q2

Your answer is correct. K = 6, so the 7<SUP>th</SUP> term has the coefficient:<O


12C6 . (-4)^6 . 3^6 = 2759049216 = 2.759049216 E09


Q3


In the expansion of (3 + 4x )^n the coefficients of x^2 and x^3 and in the ratio 3:4. Find the value of n.

You’ve done it right, just be careful doing your algebra:<O


I first expanded part of it..untill i got to x^2 and x^3...which were..
nC2 . 3^n-2 . 16x^2 (i) and nC3 . 3^n-3 . 64x^3 (ii)
I put that into the form (i)/(ii) and made it equal to 3/4.<O


To this point everything’s perfect. Let’s have a closer look to the equation (coeff. in ratio 3:4)
<O

nC2.3^n-2.16 / nC3.3^n-3.64 = 3/4


By cross multiplying we get nC2.3^n-2.64 = nC3.3^n-2.64 ( we used 3^n-3.3 = 3^n-2)
After simplifying by 64 and 3^n-2 (which is always positive), we get nC2 = nC3<O


We can find the answer immediately, using the property of combinations nCk = nCn-k (coeff. are symmetrical in Pascal’s triangle), so n = 5.
<O
Alternatively, we can use the definition of combinations: nCk = n!/k!.(n-k)!
<O</O
So nC2 = nC3 becomes: n!/2!.(n-2)!= n!/3!.(n-3)!<O

Symplify by n! (>0) and 2; then cross multiply so we get (n-2)! = 3.(n-3)!<O

We can write (n-2)! = 1.2.3……(n-3).(n-2) = (n-3)!.(n-2) and substitute in the above equation.

After simplifying by (n-3)! (which is >0), we get n-2 = 3, so n = 5.<O

You check the answer to make sure you get it right.

 

[syriangabsta]

wow thanks alot man. you rock. ^__^