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Binomial theorem1 Attachment(s) How can I prove the equation by using binomial theorem? Thanks. |

One way to start is to think $\displaystyle 2^{(n-1)} = (1 + 1)^{(n-1)} = \left (\sum_{j=0}^{n-1} \dbinom{n-1}{j} * 1^{(n-1-j)} * 1^j \right ) = \sum_{j=0}^{n-1} \dbinom{n-1}{j}.$ Now if n is even, I play around. If n = 2 $\dbinom{2-1}{0} + \dbinom{2-1}{1} = \dbinom{1}{0} + \dbinom{1}{1} = \dbinom{2 \div 2}{0} + \dbinom{2 \div 2}{1}.$ What if n = 4? $\dbinom{3}{0} + \dbinom{3}{1} + \dbinom{3}{2} + \dbinom{3}{3} = 1 + 3 + 3 + 1 = 8 =$ $1 + 6 + 1 = \dbinom{4}{0} + \dfrac{4 * 3}{2} + \dbinom{4}{4} = \dbinom{4}{0} + \dbinom{4}{2} + \dbinom{4}{4}.$ Hmm. Maybe $m \in \mathbb Z^+ \text { and } n = 2m \implies \displaystyle \sum_{j=0}^{n-1} \dbinom{n-1}{j} = \sum_{i=0}^{m}\dbinom{n}{2i}.$ Can you prove that? |

1 Attachment(s) Thank you for your hints. My prove is as follow |

$\dbinom{n}{k} =$ the number of subsets of size k in a set of size n. Since the number of even and odd subsets of a finite set are equal, we're done. There are $2^n$ subsets, and the expression we're supposed to calculate just counts the number of even subsets, which is therefore $2^{n-1}$. https://math.stackexchange.com/quest...-are-odd-sized |

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