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    This is in the syllabus isn't it?
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    I think so.
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    (Original post by dvs)
    I think so.
    Would anyone very kindly mind putting a proof up on here (induction)...I was a bit concerned I couldn't do it myself. I'm okay with de Moivre's though!

    Cheers.
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    A proof by induction for positive integer binomial powers would be.
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    emm... I can derive the binomial theorem using Maclaurin's series? Is that sufficient?
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    We know that ^{n}C_{r}=\frac{n!}{r!(n-r)!}
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    ^{n}C_{r}+^{n}C_{r-1}=\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}\\

    =\frac{n!}{r(r-1)!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)(n-r)!}\\

    =\frac{n!(n-r+1)+n!r}{r(r-1)!(n-r)!(n-r+1)}\\

    =\frac{(n+1)!}{r!(n-r+1)!\\

    =\frac{(n+1)!}{r![(n+1)-r]!}

    That is, ^{n+1}C_{r}=^{n}C_{r}+^{n}C_{r-1}

    We need to use induction to prove:
    (1+x)^{n}=^{n}C_{0}+^{n}C_{1}x+^  {n}C_{2}x^{2}+...+^{n}C_{n}x^{n}
    Assume true for n=k:
    (1+x)^{k}=^{k}C_{0}+^{k}C_{1}x+^  {k}C_{2}x^{2}+...+^{k}C_{k}x^{k}
    For n=k+1:
    (1+x)^{k+1}=(1+x)[^{k}C_{0}+^{k}C_{1}x+^{k}C_{2}x^  {2}+...+^{k}C_{k}x^{k}
    (1+x)^{k+1}=[^{k}C_{0}]+x[^{k}C_{0}+^{k}C_{1}}+x^{2}[^{k}C_{2}+^{k}C_{1}]+....+x^{k}[^{k}C_{k}+^{k}C_{k+1}]\\

(1+x)^{k+1}=^{k+1}C_{0}+^{k+1}C_  {1}x+^{k+1}C_{2}x^{2}+....+^{k+1  }C_{k+1}x^{k+1}
    Thus, if true for n=k then it is also true for n=k+1.
    Clearly it is true for n=1 as: (1+x)^{1}=1+x and by the binomial theorem^{1}C_{0}+^{1}C_{1}{x}=1+x

    Hence it is true for n=1,1+1=2,2+1=3.... and so on for all positive integral n.
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    (Original post by hello)
    Would anyone very kindly mind putting a proof up on here (induction)...I was a bit concerned I couldn't do it myself. I'm okay with de Moivre's though!

    Cheers.
    some notes and a proof attached
    Attached Images
  1. File Type: pdf binomial.pdf (99.8 KB, 124 views)
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    Cheers guys. On top form as usual!
 
 
 
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