binomial theorem: independant term of x (1 Viewer)

[wrong_turn]

thsi is from fitz,

(2x + 1/x)^2n

anyone?? test tomorrow!!!!!!

 

[acmilan]

(2x + 1/x)^2n

T_r+1
= ²ⁿC_r (2x)^r.(1/x)^2n-r
= ²ⁿC_r 2^r.x^r.x^r-2n
= ²ⁿC_r 2^r.x^{2r - 2n}

So the term independant of x has power 0, so 2r-2n = 0, r = n

The independant term is ²ⁿC_n 2ⁿ

Is that right?

 

[wrong_turn]

it was (2n)!/(n!)^2 . 2^n

im not sure what the above is :| but thanks for your answer. i see where i went wrong

 

[acmilan]

Yep thats right because

²ⁿC_n
= (2n)!/(n!(2n - n)!)
= (2n)!/(n!n!)
= (2n)!/(n!)²

 

[acmilan]

I realise this is spam, but your avatar is freaking me out.

Im assuming thats to me? If so you wouldn't be the first to say that. Criticism noted