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In a binomial expansion $(1+2)^n = \sum_{i=0}^n{n\choose i}2^{n-i}$ Why is the sum of the even $i$ 1 greater than the sum of the odd $i$?

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    @hoyland Where to go from what you've written to get the difference of 1?2012-02-19

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Just fleshing out what's already in various comments:

$\eqalign{\sum_{i{\rm\ even}}{n\choose i}2^{n-i}-\sum_{i{\rm\ odd}}{n\choose i}2^{n-i}&=\sum_{i{\rm\ even}}(-1)^i{n\choose i}2^{n-i}+\sum_{i{\rm\ odd}}(-1)^i{n\choose i}2^{n-i}\cr&=\sum_{i=0}^n(-1)^i{n\choose i}2^{n-i}=\sum_{i=0}^n(-1)^n{n\choose i}(-2)^{n-i}\cr&=(-1)^n\sum_{i=0}^n{n\choose i}(-2)^{n-i}=(-1)^n(1-2)^n=1}$