$\displaystyle\sum_{k=1}^{2n} \binom{4n}{2k}$ I thought that
=$\sum_{k=1}^{2n}\frac{4n!}{2k!(4n-2k)!}$=$\sum_{k=1}^{2n}\frac{n!}{k!(2n-k)!}$ but I don't understand what to do next, and how does the fact that k =1 and not 0 affect here?
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Have you tried some small values of n and evaluating manually? Maybe some pattern will emerge that you can try and extrapolate a general form for – Alborz Jan 11 '23 at 15:49
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Perhaps one day people will understand the difference between 'solve' and 'compute'. – ajotatxe Jan 11 '23 at 15:54
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2Hint: What is $(1-1)^{4n}$? If you do this directly vs if you do this by binomial theorem? What does this imply? How about $(1+1)^{4n}$? – JMoravitz Jan 11 '23 at 15:55
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"How does the fact that $k=1$ and not 0 affect here" Your sum is just missing the first term compared to what you would ordinarily see in other posts and so is $1$ less than the sum had we started at $k=0$ – JMoravitz Jan 11 '23 at 15:58
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So if we use the binomial theorem, I can write like this =$(1+x)^{4n}$, but it turns out that k =1 must be taken into account – le mocher Jan 11 '23 at 16:48