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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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The sum of the even is exactly the sum of the odd. Many proofs, for example put $X=-1$ in the formula of your post. –  André Nicolas Feb 19 '12 at 15:06
    
OK I changed it to X=2 case only. I guess it's not true for all N –  John Forster Feb 19 '12 at 15:20
    
The difference of the even and odd terms in the expansion of $(1+2)^n$ is the expansion of $(1-2)^n$. –  Thomas Andrews Feb 19 '12 at 15:39
    
Thanks for your answer.. why is that? –  John Forster Feb 19 '12 at 15:40
    
Rewrite as $\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k}2^{n-2k}-\binom{n}{2k+1}2^{n-2k-1}=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k}(-2)^{n-2k}+\binom{n}{2k+1}(-2)^{n-2k-1}$, since $(-1)^{2k}=1$ and $(-1)^{2k+1}=-1$. –  hoyland Feb 19 '12 at 15:49

1 Answer 1

Just fleshing out what's already in various comments:

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

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