Question on Binomial Theorem

• Deadly

  11 Sep 07, 21:13

  Find the first three terms in the expansion of (1 + 2x)^6 in ascending powers of x. Hence, or otherwise, find the coefficient of x^2 in the expansion of [(1 + 2x)^6][(2 - x)^2].
  
  Haven't really learnt Binomial Theorem yet but it's in my practice paper... thanks in advance
• CoolMyth

  11 Sep 07, 21:16

  (1+2x)^6 = 1 + (6C1)(1)(6C5)(2x) + (6C2)(1)(6C4)(2x)^2 + ...
  
  The () are all multiplication lah.
  
  For the second part, just know (2-x)^2 = 4 - 4x + x^2 and then compute the product by brute force.
• r0mE_27

  11 Sep 07, 21:20

  ans:
  
  (1+2x)^6 = 1^6 + 6C1(1^5)(2x) + 6C2(1^4)(2x)^2 + ...
  
  then later find the 2nd one
  
  (2 - x)^2 = 2 + 2C1(2^2)(-x) + 2C2(2)(-x)^2
  
  den x them. to see which will give u the power X^2
  
  6C2(1^4)(2x)^2(2) + 6C1(1^5)(2x)6C1(1^5)(2x) + 1^6(2C2(2)(-x)^2) entire thing divide by x^2 = ANSWER
• r0mE_27

  11 Sep 07, 21:21

  Originally posted by CoolMyth:

  (1+2x)^6 = 1 + (6C1)(1)(6C5)(2x) + (6C2)(1)(6C4)(2x)^2 + ...
  
  The () are all multiplication lah.
  
  For the second part, just know (2-x)^2 = 4 - 4x + x^2 and then compute the product by brute force.

  first part wrong liao la lol! (6C1)(1)(6C5)(2x) no need that one
• CoolMyth

  11 Sep 07, 21:25

  Originally posted by r0mE_27:
  
  
  first part wrong liao la lol! (6C1)(1[b])(6C5)(2x) no need that one[/b]

  Err...ok. Long time no use formula. Recalling it from memory.
• <guest>

  11 Sep 07, 21:28

  Brings me back to school days. I did binomial theorem expansion to few negative powers just to please my maths teacher to pass me
• Deadly

  11 Sep 07, 21:36

  Originally posted by r0mE_27:

  ans:
  
  (1+2x)^6 = 1^6 + 6C1(1^5)(2x) + 6C2(1^4)(2x)^2 + ...
  
  then later find the 2nd one
  
  (2 - x)^2 = 2 + 2C1(2^2)(-x) + 2C2(2)(-x)^2
  
  den x them. to see which will give u the power X^2
  
  6C2(1^4)(2x)^2(2) + 6C1(1^5)(2x)6C1(1^5)(2x) + 1^6(2C2(2)(-x)^2) entire thing divide by x^2 = ANSWER

  um sorry didn't really get it. can explain in simpler terms cuz I havent really been exposed to this topic yet.
• r0mE_27

  11 Sep 07, 22:57

  the last part.
  
  u needa get
  
  1 times x^2 to get x^2
  x times x to get x^2
  x^2 times 1 to get x^2
  
  hence u just cross multiply lo, those that get u x^2 and den just add them together
• striderz_o405

  11 Sep 07, 22:59

  refer to the formulae sheet.
• wishboy

  13 Sep 07, 23:54

  binomial theorem only got a few formula
  
  for (1+x)^n,
  T(r+1) = (nCr) * (x^r)
  
  for (y-x)^n,
  T(r+1) = (nCr) * [y^(n-r)] * (x^r)
  
  using these 2 formula, u can find a specific term by substituting the number of the term into T which is r+1.
  then r would be the number of the term minus 1
  other values sub in oso to find that term
  
  u oso can use them to find for x or y to the power of any number, uhh this one i dunno how to explain very well so i think nvm ba, let ur teacher teach u better
• ^tamago^

  13 Sep 07, 23:56

  Originally posted by Deadly:

  um sorry didn't really get it. can explain in simpler terms cuz I havent really been exposed to this topic yet.

  ok first do u noe how to do a single binomal expansion?
• Deadly

  14 Sep 07, 05:55

  its okay la i managed to do the question with the help rendered in this thread already. thanks ppl
• eagle

  14 Sep 07, 06:58

  Originally posted by Deadly:

  its okay la i managed to do the question with the help rendered in this thread already. thanks ppl

  
• r0mE_27

  14 Sep 07, 12:47

  you are welcome
• wishboy

  14 Sep 07, 19:06

  Originally posted by Deadly:

  its okay la i managed to do the question with the help rendered in this thread already. thanks ppl

  my pleasure...