Binaomial Expansion Nov 27 09
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Binaomial Expansion Nov 27 09

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Binaomial Expansion Nov 27 09 Binaomial Expansion Nov 27 09 Presentation Transcript

  • The Binomial  Theorem,  Fibonacci, and  Vitruvian Dan Vitruvian Dan by flickr  user nogoodreason
  • Multiply each of the following (x + y)0 = (x + y)1 = (x + y)2 =  (x + y)3 = (x + y)4 =  (x + y)5 =
  • What's the pattern?  1 1     1 1     2     1 1     3     3     1 1     4     6     4     1
  • 1. The sum of the exponents is always equal to     the power of the binomial.  2. The exponent of the first term begins with the     same value as the power of the binomial and     decreases by one in each succesive term.  3. The exponent of the second term appears in the     second term of the expansion and increases by     one until it matches the power of the binomial. 
  • 4. The number of terms is one more than the      power of the binomial.  5. The coefficients are the combinations of the     power number beginning with C(n, 0) and     ending at C(n, n) and are symmetrical.  n 0 n­2 2 (x + y)n = nC0 x y   + nC1 xn­1y1  + nC2 x y   + . . . . .  + nCn x0yn
  • Evaluate each term ...
  • (x + y)7 =
  • (2x ­ y)4 = 
  • Any individual term, let's say the ith term, in a  binomial expansion can be represented like this: n ­ (i ­ 1) (i ­ 1) ti = nC(i ­ 1)a b Find the 4th term in the expansion of (2 + x)7
  • Find the 5th term 8 ( x2 ­ 2 ) x
  • Determine the indicated term in each expansion. the 8th term in the expansion of  10 (x ­ 2 )
  • Find the term that contains x7 in the expansion of 9 (x3 ­ 1 x )
  • Find the term that contains x18 in the expansion of  10 ( x3 ­ 1 ) x
  • Try some more Exercise 34, questions 1 ­ 8
  • Attachments test by flickr user foreversouls Vitruvian Dan by flickr user nogoodreason