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fpmaker

  • 2 years ago

Binomial Expansion: (n choose k) + (n choose (k+1))...need to use the factorial form to go through to step and eventually prove the answer... = ((n+1) choose (k+1))...

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  1. across
    • 2 years ago
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    \[\begin{align} \binom nk+\binom n{k+1}&=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}\\ &=\frac{1}{n-k}\left[\frac{n!(k+1)}{(k+1)!(n-k-1)!}+\frac{n!(n-k)}{(k+1)!(n-k-1)!}\right]\\ &=\frac{(n+1)!}{(k+1)!(n-k)!}=\binom{n+1}{k+1}. \end{align}\]

  2. fpmaker
    • 2 years ago
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    Thank you! My algebra with factorials is a little weak..is there any way you could briefly explain how you got to each step? Thanks again!!

  3. across
    • 2 years ago
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    The only algebraic manipulation that's worth mentioning is that \(n!=n(n-1)!\). Both the LHS and the RHS on first line are self-explanatory. On the second line, I factored out \((n-k)^{-1}\) from both terms (notice I multiplied the second term by \(n-k\)) to equate denominators (using the idea I just mentioned). Finally, I just expanded and added the numerators on the third line to prove the equality.

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