small binomial q. (1 Viewer)

[shkspeare]

⁵C₂ + ⁵C₃ = ⁶C₃

ok i know they both equal to 20

how does the LHS simplify to give RHS

 

[Supra]

5x4
----- = 10
2x1

5x4x3
-------- = 10
3x2x1

add the 2 up and u get 20

 

[Supra]

sorr y it dont look clear on the post but i cant help it

 

[shkspeare]

yes i know how to get 20

i dont get how the books answer simplifies it to ⁶C₃ on the next line tho...

 

[shkspeare]

oops

wot i meant was
the answer is 20.. yes i can get that

but wut i dont get is the books working out saying that it simplifies to 6C3 which simplifies to 20

 

[evilc]

hope this explains it

 

[Grey Council]

nice evilc.

However, I would have changed one thing in your working...

You should have written "Viola!!" when you wrote LHS = RHS.

heehehe, nice work.

 

[abdooooo!!!]

Originally posted by shkspeare
yes i know how to get 20

i dont get how the books answer simplifies it to ⁶C₃ on the next line tho...

hey shkspeare i get what you mean... because there is a formula which looks like this: ⁿC_k = ^n-1C_k-1 + ^n-1C_k which what ever book you use has used it to derive ⁶C₃. i think you'll learn how to prove this in 2/3u course by the properties of the coefficients of binomial theorem.

so just read your books

 

[shkspeare]

ahhh thx evilc....

abdoo... u cleared everything up

 

[Supra]

ohhh i now get wot ur talking bout...lol bit fukn l8 tho

that factorial stuff in abdoo's solution is exactly the same as jus doing (5x4)/(1x2)

i forgot to do the 6c3 tho, sorry

 

[KeypadSDM]

R.T.P. ⁵C₂ + ⁵C₃ = ⁶C₃

⁵C₂ + ⁵C₃
=(5!)/(2!3!) + (5!)/(3!2!)
=(2)(5!)/(2!3!)
=(5!)/(3!)
=(6*5!)/(6*3!)
=(6!)/(3!3!)
=⁶C₃

A slightly quicker method.

 

[shkspeare]

ohhh thx keypad

ummm just a few more questions >=\

1)¹⁰C_k / ¹⁰C_k-1

2)Simplify (sqrt(2) + 1)⁴ +(sqrt(2) - 1)⁴

3^The expansion of (1 + ax)ⁿ is given by 1 - 24x + 252x² - ...

Find a and n

bah the first one i keep getting 11/k
and the second one im wondering if theres a shortcut method
the third one im not quite sure =\

also are we allowed to assume that ^n-kC_n-1-k = n - k ?

 

[shkspeare]

anyone?

 

[wogboy]

1. Use the nCk = n!/k!(n-k)! formula.

10Ck / 10C(k-1)
= { 10!/k!(10-k)! } / { 10!/(k-1)!(11-k)! }
= (k-1)! * (11-k)! / (k! * (10-k)! )
= (k-1)!/k! * (11-k)!/(10-k)!
= 1/k * (11-k)
= 11/k - 1

2. Try squaring (sqrt(2) + 1) and (sqrt(2) - 1) each twice, then add together. Should be quicker than using the full bionomial expansion.

(sqrt(2) + 1)^2
= 2 + 2*sqrt(2) + 1
= 3 + 2*sqrt(2)
likewise,
(sqrt(2) - 1)^2
= 3 - 2*sqrt(2)

squaring them both again,
(sqrt(2) + 1)^4
= 9 + 12*sqrt(2) + 8
= 17 + 12*sqrt(2)
likewise,
(sqrt(2) - 1)^4
= 17 - 12*sqrt(2)

adding them together.

(sqrt(2) + 1)^4 + (sqrt(2) - 1)^4
= 34

If you do 4U, you'll notice the resemblance of surds to complex numbers, from this question. Just as a complex no. has real and imaginary components, a surd has a rational and irrational components, and they act quite similarly.

e.g. (sqrt(2) + i)^4 + (sqrt(2) - i)^4
= 34 too.

3.

(1 + ax)^n = 1 - 24x + 252x^2 - ...

generally,
(1 + ax)^n = 1 - n*ax + (nC2)*a^2*x^2 - ...

comparing the two equations,
na = -24
nC2 * a^2 = 252

but nC2 =n!/2!(n-2)! = n(n-1) /2

a^2 * (n^2 - n) = 504
576/n^2 * (n^2 - n) = 504
576 - 576/n = 504
576/n = 72

n = 8
a = -3

 

[shkspeare]

oHHH!! i get it now... thanks!!

 

[victorling]

Originally posted by shkspeare
⁵C₂ + ⁵C₃ = ⁶C₃

ok i know they both equal to 20

how does the LHS simplify to give RHS

construct the pascal triangle
and you will get the clue

 

[KeypadSDM]

Re: Re: small binomial q.

Originally posted by victorling
construct the pascal triangle
and you will get the clue

Actually that doesn't help you if you want to do it algebraicly.