Binomial Theorem (1 Viewer)

[vds700]

it's the number of ways of choosing 2 from each set of parallel lines, isn't it?
6C2 * 9C2 = 540

thanks man you're a genius!

 

[lyounamu]

thanks man you're a genius!

Yeah, that question was confusing. I ended up spending 5 minutes before I realised that it was a combination question.

 

[lyounamu]

I think i might have figured (iii) out. You expand it so you get

sum(0 -> n ) (r+1)nCr= sum(0 -> n) r nCr + nCr = n.2^(n-1) +2^n.

And can someone have a look at the last question on that page please. There must be a better way to do it than counting, which i really can't be bothered doing.

I don't think that you expand the question for that. I think you actually have to find a particular expression that describes the sum. But in saying that, I didn't really understand the question.

 

[lolokay]

I don't think that you expand the question for that. I think you actually have to find a particular expression that describes the sum. But in saying that, I didn't really understand the question.

I don't really think there's anything wrong with expanding it. I would imagine that that's how it's meant to be done (considering you've found the expression for each part already)

 

[vds700]

I don't really think there's anything wrong with expanding it. I would imagine that that's how it's meant to be done (considering you've found the expression for each part already)

In the solution they do it slightly differently, but they get the same answer.

They multiply (1 + x)^n = nC0 + nC1x +... by x and differentiate, then let x = 1.

 

[FDownes]

I've got another binomial question that I'm stuck on. It asks;

The coefficients of x⁴ and x⁵ in the expansion of (3 - x)ⁿ are equal in magnitude but opposite in sign. Find the value of n.

 

[lolokay]

the coefficients are given by
nCr 3^n-r

so:
n!/4!(n-4)! 3^n-4 = n!/5!(n-5)! 3^n-5
simplifying;
3/(n-4) = 1/5
n=19

 

[lyounamu]

I've got another binomial question that I'm stuck on. It asks;

The coefficients of x⁴ and x⁵ in the expansion of (3 - x)ⁿ are equal in magnitude but opposite in sign. Find the value of n.

x^4: nC4 . 3^(n-4) . (-x)^4

x^5: nC5 . 3^(n-5) . (-x)^5

And since they are opposite in signs (but (-x)^5 is always negative with (-x)^4 always positive):

nC4 . 3^(n-4) . (-x)^4 = -nC5 . 3^(n-5) . (-x)^5
3 . nC4 = x nC5
3n!/(n-4)!4! = x . n!/(n-5)!5!
3n! . (n-5)! . 5! = x . n! . (n-4)! . 4!
3(n-5)! . 5! = x . (n-5)! . (n-4) . 4!
3 . 5! = x .(n-4) 4!
3 . 5 = x(n-4)

But we only take the coefficeints so:

So 15 = n -4
n = 19

 

[FDownes]

nC4 . 3^(n-4) . (-x)^4 = -nC5 . 3^(n-5) . (-x)^5
3 . nC4 = x nC5
3n!/(n-4)!4! = x . n!/(n-5)!5!
3n! . (n-5)! . 5! = x . n! . (n-4)! . 4!
3(n-5)! . 5! = x . (n-5)! . (n-4) . 4!
3 . 5! = x .(n-4) 4!
3 . 5 = x(n-4)

I'm afraid I don't quite follow the lines of working in bold. Where does the (n - 5)! on the right of the fifth line come from? And why does the 4! on the right of the sixth line disappear?

 

[lyounamu]

I'm afraid I don't quite follow the lines of working in bold. Where does the (n - 5)! on the right of the fifth line come from? And why does the 4! on the right of the sixth line disappear?

Sorry, I may have skipped few lines along the way.

(n-5)! come from (n-4)!. Recall: (n-4)! = (n-5)! . (n-4)

And 5! = 4! . 5

 

[FDownes]

Ah, I get it now. Thanks for that.

 

[FDownes]

Okay, new question;

a) Simplify ⁸C_k/⁸C_{k - 1}.

[This part is relatively easy, it simplifies to (9 - k)/k].

b) Hence or otherwise, find the greatest coefficient of the expansion of (1 + x)⁸.

[This part I'm having trouble with...]

Can anyone please help me?

 

[lolokay]

the greatest coefficient will occur at the maximum value for which 8Ck>8C(k-1)
so 8Ck/8C[k - 1] > 1
(9-k)/k>1
9>2k
k<4.5
so the greatest value of k is 4
8!/4!4! = 70

 

[FDownes]

Thanks for that.

I seem to be stuck on yet another question, however. I'm not sure what I did wrong, but I just can't seem to get the right answer out. It asks;

Find the greatest coefficient of (2x - 1)⁵.

It should be fairly simple. The answer is apparently +/-80.

 

[lolokay]

the coefficient of the rth term is given by (I'm letting the first coefficient be r=0)
(-1)^r2^5-r 5!/r!(5-r)!
we want to find the lowest value of the coefficient such that it is greater in magnitude than the coefficient after it (or however you like to set it up. You can find the greatest coefficient such that it is greater than the coefficient before it)
ie.
2^5-r 5!/r!(5-r)! > 2^4-r 5!/(r+1)!(4-r)!
which you simplify to:
2/(5-r) > 1/(r+1)
2r + 2 > 5 - r
r > 1
Since it comes out to be an integer, I think you can assume the the coefficients for 2 (the lowest number to make the inequality true) and 1 (if you replace the > with >=) will have the same coefficient.

if we substitute r=1,2 into the formula (-1)^r2^5-r 5!/r!(5-r)!
we get -+80.
(-1)^r2^5-r 5!/r!(5-r)!

 

[FDownes]

Thanks again.