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C2-Binomial expansion

In the binomial expansion of (2k+x)^n, where k is a constant & n is a positive integer, the coefficient of x^2 is equal to the coefficient of x^3.
Prove that n = 6k +2

How can we prove this? Please help me.

Reply 1

geton

thirdrate

x^2 term = nC2 (2k)^(n-2) x^2
x^2 coeffiecient = nC2 (2k)^(n-2)

Thank you

Reply 2

jasonlu

I get what you guys are saying but how do you prove it? You have
2k^n-2 = 2k^n-3
Then how do you find n? All i can fink about is taking logs..

Reply 3

icapturethecastle

i think u do have to take logs

Reply 4

icapturethecastle

so u do or don't take logs?
:s-smilie:

You don't.

ooh i see...

one day i'll conquer binomial expansion :s-smilie:

Reply 7

Narev

I'll try

(2k + x)^n

= (2k)^n (1 + x/2k)^n

Consider the first few terms of the expansion of (1 + x/2k)^n using binomial theorem.
(We can ignore (2k)^n as it'll be multiplied to the coefficient of the x^2 and x^3 term in (1 + x/2k)^n )

1 + nC1 (x/2k) + nC2 (x/2k)^2 + nC3 (x/2k)^3 + ...

Given coefficient of x^2 = coefficient of x^3

nC2 / (2k)^2 = nC3 / (2k)^3

n!/ {2! . (n-2)! . 4k^2} = n! / {3! . (n-3)! . 8k^3}

8(n-2)! (k^2) = 48 (n-3)!(k^3)

To make it a bit clearer

8(1)(2)(3)..(n-3)(n-2) (k^2) = 48 (n-3)! (k^3)

8(n-2)(k^2) = 48k^3
8(n-2)=48k
n-2 = 6k
n = 6k + 2

QED

Reply 8

Spiderwoman

Thanks alot Narev for the help :biggrin: