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  1. Show that$$(1+x)^n=\binom{n}{0}+\binom{n}{1} x+\binom{n}{2} x^2+\ldots+\binom{n}{n-1} x^{n-1}+\binom{n}{n} x^n.$$
  2. Hence deduce that:
    1. \(\displaystyle \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\ldots+\binom{n}{n-1}+\binom{n}{n}=2^n\)
    2. \(\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\ldots+(-1)^n\binom{n}{n}=0\)

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