Binomial Expansion

• Divanhot

  6 Jun 09, 02:05

  Find the term independent of x in the expansion of (2x^3 - 1/4x^2) ^10

  http://img512.imageshack.us/i/binomial.jpg/

  attached is the question

  i used

  Tr+1= 10cr (2x^3) ^10-r (-1/4x^2)^r

  = 10cr 2x^30-3r(-1/4)^r(x^-2r)

  = 10cr 2(-1/4)^r x^30-5r

  i equate 30-5r=0, r=6

   

  subst r-6 in 10c6 * 2(-1/4)^6, but answer is wrong. correct ans is 105/128

  i suspect is i shouldnt place the 2 in front of (-1/4)^r, do advise

• Freddie365

  6 Jun 09, 04:47

  Hi im getting a problem to solving this .  Who were giving  GCE A2 LEVEL'S ACCOUNTING

   

• wee_ws

  6 Jun 09, 08:00

  Hi,

    Tr+1= 10 C r (2x^3)^(10 - r) (-1/4x^2)^r

    = 10 C r (2)^(10 - r) (-2)^(-2r) (x)^(30 - 5r).

    Let 30 - 5r = 0, we obtain r = 6.

    Answer of 105/128 is correct. Thanks!

  Cheers,
  Wen Shih