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 $i$?
Sum of even binomial expansions 1 more than sum of odd terms
0
$\begingroup$
combinatorics
-
0@hoyland Where to go from what you've written to get the difference of 1? – 2012-02-19
1 Answers
1
Just fleshing out what's already in various comments:
$\eqalign{\sum_{i{\rm\ even}}{n\choose i}2^{n-i}-\sum_{i{\rm\ odd}}{n\choose i}2^{n-i}&=\sum_{i{\rm\ even}}(-1)^i{n\choose i}2^{n-i}+\sum_{i{\rm\ odd}}(-1)^i{n\choose i}2^{n-i}\cr&=\sum_{i=0}^n(-1)^i{n\choose i}2^{n-i}=\sum_{i=0}^n(-1)^n{n\choose i}(-2)^{n-i}\cr&=(-1)^n\sum_{i=0}^n{n\choose i}(-2)^{n-i}=(-1)^n(1-2)^n=1}$