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
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combinatorics
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1The sum of the even is exactly the sum of the odd. Many proofs, for example put $X=-1$ in the formula of your post. – 2012-02-19
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0OK I changed it to X=2 case only. I guess it's not true for all N – 2012-02-19
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0The difference of the even and odd terms in the expansion of $(1+2)^n$ is the expansion of $(1-2)^n$. – 2012-02-19
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0Thanks for your answer.. why is that? – 2012-02-19
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0Rewrite 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 $(-1)^{2k+1}=-1$. – 2012-02-19
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0@John Forster: Before you edited your post, I thought that you were noticing that the sum of the coefficients of the even powers of $x$ (including the constant term $x^0$) is the same as the sum of the coefficients of odd powers of $x$. For example, look at $(1+x)^4=1+4x+6^2+4x^3+x^4$. We have $1+6+1=4+4$. Same is true for $(1+x)^n$. – 2012-02-19
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0@hoyland Where to go from what you've written to get the difference of 1? – 2012-02-19