Binomial Theorem (1 Viewer)

[FDownes]

I'm not sure how to approach this question, can anyone help me? It asks;

Simplify;
a) ^{n - 1}C_{k - 1} + ^{n - 1}C_k

b) ¹⁰C_k ÷ ¹⁰C_{k - 1}

 

[vds700]

I'm not sure how to approach this question, can anyone help me? It asks;

Simplify;
a) ^{n - 1}C_{k - 1} + ^{n - 1}C_k

b) ¹⁰C_k ÷ ¹⁰C_{k - 1}

heres my solution

 

[lyounamu]

heres my solution

That looks like mine. (mine is more factorised though)

 

[FDownes]

I got a similar answer for part a), but according to the textbook the answer should be ⁿC_k. The answer for part b) is definately correct though.

 

[lyounamu]

I got a similar answer for part a), but according to the textbook the answer should be ⁿC_k. The answer for part b) is definately correct though.

vds made a mistake in the fourth line. It should be just k not k!.

Then you get (n-1)!(k+n-k)/(n-k)!k! = n!/(n-k)!k! = nCk

 

[vds700]

I got a similar answer for part a), but according to the textbook the answer should be ⁿC_k. The answer for part b) is definately correct though.

part a) is actually the second pascal triangle relation, look in your textbook, should have a proof

 

[vds700]

vds made a mistake in the fourth line. It should be just k not k!.

Then you get (n-1)!(k+n-k)/(n-k)!k! = n!/(n-k)!k! = nCk

oops sorry- good work namu

 

[vds700]

got a q i need help with, part b).

Thanks

 

[lyounamu]

got a q i need help with, part b).

Thanks

bi) When x=-1,

(1+-1)^n = nCo + nC1 . (-1)^1 + ... + nC(n-1) . (-1)^(n-1) + nCn . (-1)^n
0 = nCo - nC1 + nC2 + ... - nC(n-1) + nCn
Move all the minus things to the other side so you get:
nC1 + nC3 + .... + nC(n-1) = nCo + nC2 + nC4 + ...+ nCn
So P=Q where P is nC1 + nC3+...+nC(n-1) and Q = nC0 + nC2 + ... +nCn
Now when x = 1,

(1+1)^n = nCo + nC1 + nC2+ nC3 +...+nCn

2^n = nCo + nC1 + nC2 + nC3 +... +nCn

So 2^n = 2P since ( P = Q)

So 2^n/2 = P
= nC1 + nC3 + .... + nC(n-1)

ii) (1+x)^n = nC0 + nC1 . x + ... + nCn . x^n
Differentiate both sides:

n(1+x)^n-1 = nC1 + 2 nC2 . x + ... + n nCn . x^n-1
When x = 1, n2^(n-1) = nC1 + 2nC2 + ...n . nCn = the result given.

iii) I don't know really know at this stage but I will try,

 

[FDownes]

Thanks, guys. Got time for another?

The expansion of (1 + ax)ⁿ is given by 1 - 24x + 252x² - ... . Find values for a and n.

 

[lyounamu]

Thanks, guys. Got time for another?

The expansion of (1 + ax)ⁿ is given by 1 - 24x + 252x² - ... . Find values for a and n.

x: nC1 . 1^(n-1) . (ax)^1 = -24x
So anx = -24x
an = -24 ...(1)

x^2: nC2 . 1^(n-2) . (ax)^2 = nC2 . a^2 . x^2 = 252x^2
So nC2 . a^2 = 252 ...(2)

From (1), a^2n^2 = 576

Divide that by (2) you get:

16/7 = n^2/nC2

After simplifying you get: n/(n-1) = 8/7
So n = 8
So a = -3

 

[FDownes]

Ah, that makes sense. I have no idea why I didn't realise that ⁿC₁ = n, and that 1^{n - 1} = 1. Thanks.

 

[vds700]

bi) When x=-1,

(1+-1)^n = nCo + nC1 . (-1)^1 + ... + nC(n-1) . (-1)^(n-1) + nCn . (-1)^n
0 = nCo - nC1 + nC2 + ... - nC(n-1) + nCn
Move all the minus things to the other side so you get:
nC1 + nC3 + .... + nC(n-1) = nCo + nC2 + nC4 + ...+ nCn
So P=Q where P is nC1 + nC3+...+nC(n-1) and Q = nC0 + nC2 + ... +nCn
Now when x = 1,

(1+1)^n = nCo + nC1 + nC2+ nC3 +...+nCn

2^n = nCo + nC1 + nC2 + nC3 +... +nCn

So 2^n = 2P since ( P = Q)

So 2^n/2 = P
= nC1 + nC3 + .... + nC(n-1)

ii) (1+x)^n = nC0 + nC1 . x + ... + nCn . x^n
Differentiate both sides:

n(1+x)^n-1 = nC1 + 2 nC2 . x + ... + n nCn . x^n-1
When x = 1, n2^(n-1) = nC1 + 2nC2 + ...n . nCn = the result given.

iii) I don't know really know at this stage but I will try,

thanks heaps man. I'll see if i can get the solution off mr hamam on monday for part (iii)

 

[wendus]

fucking hate binomials

 

[lyounamu]

thanks heaps man. I'll see if i can get the solution off mr hamam on monday for part (iii)

Yeah, he should have one. I don't get part (iii). I tried to relate that one to the part ii but I failed. I better look at that question closely.

 

[lyounamu]

fucking hate binomials

It's okay. It's much better than Perm & Comb.

 

[wendus]

AGREED. pointless useless afdsafadfewt

 

[addikaye03]

It's okay. It's much better than Perm & Comb.

agreed, perms & combs are devils work. I kinda like binomials

 

[vds700]

Yeah, he should have one. I don't get part (iii). I tried to relate that one to the part ii but I failed. I better look at that question closely.

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.

 

[lolokay]

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.

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