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Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $ is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;

Let $a=1, b=-1$ and $n=1$ ,$k=0$ , using Binomial expression.

$$(a+b)^n= \sum \binom{n}{k} a^{k}b^{n-k}$$

Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$

Thus :For all $n\geq1: \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$ .

Michael Hardy
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user142943
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  • Please typeset this. – Batman May 24 '14 at 05:03
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    A proof that the result is true can be produced using ideas contained in what you wrote. Note that $1+(-1)=0$. – André Nicolas May 24 '14 at 05:03
  • @user142943: You are almost using TeX correctly, but leaving out the $$$ signs around TeX expressions. Please see how your previous post was edited, or this one once somebody does it. – André Nicolas May 24 '14 at 05:06
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    It looks like you're saying that the two sums are unequal for all $n\ge1$? Then how come for $n=3$ I get $\binom 60+\binom62+\binom64+\binom66=1+15+15+1=32$ and $\binom61+\binom63+\binom65=6+20+6=32$? – bof May 24 '14 at 05:12

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$$\displaystyle(1-1)^{2n}=0=\{\binom{2n}{0}+\binom{2n}{2}...\}-\{\binom{2n}{1}+\binom{2n}{3}...\}$$

$$\implies \{\binom{2n}{0}+\binom{2n}{2}...\}=\{\binom{2n}{1}+\binom{2n}{3}...\}$$

evil999man
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