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C2 HEELP

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T.Q:

--- Quote from: XFlufferzX on January 06, 2010, 07:13:06 pm ---Incase if my handwriting wasnt clear there ya go :
Using the identity:
(1+kx)^n = 

1 + n (kx) + n (kx) + n(n-1)/2! k^2 + n(n-1)(n-2)/3! kx^3

1+ Ax + Bx^2 + Bx^3

so A = nk
    B=  n(n-1)/2! k^2
    B= n(n-1)(n-2)/3! k^3

(2!= 2 , 3! =6)

Equating the problem:

n(n-1)/2 k^2 = n(n-1)(n-2)/6 k^3

Simplify it and you'll get

3= (n-2) k

For b)

3 = (n-2) k
Simplify

3 = nk -2k

3 = A -2k

3 = 4 - 2k

K = 1/2

Solving for n

A = nk

4 = n(0.5)

n=8

Its a good question, from which paper did you get dat?

--- End quote ---
the paper is attached

XFlufferzX:
thanks!  :D

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