Author Topic: Pure Math - Series (AS)  (Read 720 times)

Offline Diablo

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Pure Math - Series (AS)
« on: January 28, 2010, 07:25:18 pm »
Hola!

Help will be appreciated with the following question:

If the first three terms in the expansion of (1 + kx)n in ascending powers of x are 1-6x+33/2x2 , find the values of k and n.

I'm new to this chapter and the ncr stuff, so I can't really apply Algebra in such questions :D.

Thanks!

Offline Saladin

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Re: Pure Math - Series (AS)
« Reply #1 on: January 29, 2010, 12:42:39 pm »
you'r first 3 terms are confusing. Please clearly state the first 3 terms.

Offline astarmathsandphysics

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Re: Pure Math - Series (AS)
« Reply #2 on: January 29, 2010, 12:51:44 pm »
One mo

Offline astarmathsandphysics

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Re: Pure Math - Series (AS)
« Reply #3 on: January 29, 2010, 01:04:41 pm »
(1 + kx)n in ascending powers of x are 1-6x+33/2x2 , find the values of k and n.
(1 + kx)n=1+nkx+n(n-1)/2 k^2 x^2=1-6x+33/2x^2
nk=6 andn(n-1)/2k^2=33/2
kn=6 so k^2=36/n^2 and n(n-1)k^2=33 so n(n-1)*36/n^2==36-36/n=33 so 3=36/n and n=12 and thenk=1/2

Offline Diablo

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Re: Pure Math - Series (AS)
« Reply #4 on: January 29, 2010, 10:28:59 pm »
Thanks for your replies. I managed to solve it earlier today after going through some examples.

Thanks Again  :D