AS MATHS Help Needed P1

[XPFMember]

Assalamoalaikum!!

Well...i have my exam within 3-4 days...and i have a fw doubts...which i'll be posting soon...Anybody ready to help?

anyways first question is here...ch: series

given that first 3 terms of (1 + ax)^n = 1 + 24x + 270x^2
find the values of a and n....dont know how to solve...can anybody help?? please..
Jazak AllahKhair!

 

[usman]

(1 + ax)^n = (1)^n + n(1)^(n - 1)*(ax) + (1/2)*n(n - 1)(1)^(n - 2)*(ax)^2
= 1 + 24x + 270x^2
Now, comparing coefficients ---->
an = 24 .....[1] (1/2)*n(n - 1)*a^2 = 270 .....[2]
Solving them simultaneously yields n = 16 and a = 3/2 = 1.5

 

[XPFMember]

frm where did u get 1/2 n(n-1)

the way we learnt is nc0..and all that!!

 

[Nibz]

U have: (1 + ax)^n => 1 + nax + (n (n - 1))/(2 factorial) ax^2 = 1 + 24x + 270 x^2
Compare now => na = 24 so a = 24/n
=> [ (n^2 - n) (a)^2 ] /2 = 270
=> 576n^2 - 576n = 540n^2
=> n^2 - 16n = 0
n = 16
a = 24/16 = 1.5

P.S here is the formula for Binomial Expansion:
(1 + x)^n = 1 + nx + [ n ( n -1) ]/2 (factorial) x^2 + [ n (n-1) (n-2) ]/3(factorial) x^3 ...........and so on!!!

 

[XPFMember]

whats factorial..and it'd be better if u attach a word doc here..n write the eqn in fraction form..plz

 

[Nibz]

Whats so difficult in this??
Which equation doesnt ring a bell?
Factorial (!) 2! = 2x1
3! = 3x2x1
4! = 4x3x2x1
5! = 5x4x3x2x1 and so on and so forth!!

 

[XPFMember]

oh...actually we didnt do this...'' ! '' thing..we took it as nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2...and so on...

 

[Nibz]

Can u apply that formula here? :|

 

[XPFMember]

using that i wasnt able to do it..!!

 

[Nibz]

That is why I used that formula!

 

[XPFMember]

but i didnt know that formula i'll check that in the text book...i hope i'll understand it then Insha Allah

 

[Nibz]

U must know it!