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Bionomial expansion

  1. Nov 25, 2007 #1
    hi i have a whole page of bionomial expansion homework from as teacher and have never encountered it before here is an example of one of my questions could someone please show me how its done.

    example:
    find the coefficient of x (cubed) in the expansion of (2 + 3x) (to the power of 5)

    i am completly stuck,

    thanks in advance.
     
  2. jcsd
  3. Nov 25, 2007 #2
    x (cubed) in the expansion of (2 + 3x) (to the power of 5)

    well thats kinda easy actually, just think of
    (2 + 3x)[tex]^{5}[/tex] as (2 +3x)(2 +3x)(2 +3x)(2 +3x)(2 +3x)
    so just multiply out using foil....looks like its gonna take a while but im assuming its just for practice and to get u into the mood of foil for other questions about functions
     
  4. Nov 25, 2007 #3
    To solve (a+b)[tex]^{n}[/tex], Pascal's triangle may help you:
    1 n=0
    1 1 n=1
    1 2 1 n=2
    1 3 3 1 n=3
    1 4 6 4 1 n=4
    1 5 10 10 5 1 n=5
    1 6 15 20 15 6 1 n=6
    1 7 21 35 35 21 7 1 n=7
    1 8 28 56 70 56 28 8 1 n=8

    I've shown the first 9 lines, however, you can continue writing lines indefinitely. Every number is the sum of upper two. For example in n=3 line, 1 = 0 + 1, 2 = 1 + 1, 1 = 1 + 0. In n=4 line 1 = 0 + 1, 3 = 1 + 2, 3 = 2 + 1, 1 = 1 + 0 and so on.

    Numbers in these lines coefficients for expansion of (a+b)[tex]^{n}[/tex].

    (a+b)[tex]^{n}[/tex] can be expanded to Ca[tex]^{n}[/tex]+Ca[tex]^{n-1}[/tex]b+Ca[tex]^{n-2}[/tex]b[tex]^{2}[/tex]+...+Ca[tex]^{2}[/tex]b[tex]^{n-2}[/tex]+Cab[tex]^{n-1}[/tex]+Cb[tex]^{n}[/tex] where C are the coefficients from line n.

    For example:
    (a+b)[tex]^{3}[/tex]=a[tex]^{3}[/tex]+3a[tex]^{2}[/tex]b+3ab[tex]^{2}[/tex]+b[tex]^{3}[/tex]

    To expand (2+3x)[tex]^{5}[/tex] we must take the coefficients from n=5 line =>
    (2+3x)[tex]^{5}[/tex]=2[tex]^{5}[/tex]+5*2[tex]^{4}[/tex]*3x+10*2[tex]^{3}[/tex]*(3x)[tex]^{2}[/tex]+10*2[tex]^{2}[/tex]*(3x)[tex]^{3}[/tex]+5*2*(3x)[tex]^{4}[/tex]+(3x)[tex]^{5}[/tex] = 32+270x+720x[tex]^{2}[/tex]+1080x[tex]^{3}[/tex]+810x[tex]^{4}[/tex]+243x[tex]^{5}[/tex]

    If I calculated correctly, than from this you can see that the coefficient of x cubed is 1080. I think that expanding the binomial this way is much easier than multiplying.

    I hope that helps.

    P.S. If you see the triangle with a straight angle, than look here.
     
  5. Nov 25, 2007 #4
    You can get at this quicker through binominal coefficients: (3X+2)^5, the coefficient on x^3 will be 5!/(3!2!) = 10, but you have to consider the three on x and the 2 as well.
     
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