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### Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots$

Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots$$ I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a ...
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Consider the identity $$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$ where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity $$\sum_{k=0}^n(-... 4answers 2k views ### Show that \sum_{k=0}^n\binom{3n}{3k}=\frac{8^n+2(-1)^n}{3} The other day a friend of mine showed me this sum: \sum_{k=0}^n\binom{3n}{3k}. To find the explicit formula I plugged it into mathematica and got \frac{8^n+2(-1)^n}{3}. I am curious as to how one ... 4answers 349 views ### Find the value of \binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000} Find the value of \binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000} I've seen many complex proofs. I am looking for an elementary proof. I know the fact that \binom{... 2answers 9k views ### Number of even and odd subsets [duplicate] Suppose we have the following two identities: \displaystyle \sum_{k=0}^{n} \binom{n}{k} = 2^n \displaystyle \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0 The first says that the number of subsets of ... 3answers 932 views ### Evaluating Combination Sum \sum{n+k\choose 2k} 2^{n-k} Evaluate$$\sum_{k=0}^n{n+k\choose 2k} 2^{n-k}$$So im not really sure how to begin with this. I would imagine we start with dividing out 2^{n}, but not really sure much past that 2answers 800 views ### lacunary sum of binomial coefficients I am sure there must be a known estimate for sums of the form:$$ S_d(n)=\sum_{j=0}^n \binom{dn}{dj} $$where d\ge 1. For d=1, the answer is obvious. For d=2, the answer is here: Sum with ... 1answer 2k views ### Sum of every kth binomial coefficient. It is widely known that$$\sum_{m=0}^{n} {n\choose m} = 2^n$$and that$$\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose 2m} = 2^{n-1} Both results can be proven by exploting the nature of the roots ...
The complete question I would like to answer is: Given positive integers $k,n$, how many descending lists of non-negative integers $(x_1~x_2\ldots x_k)$ are there such that $\sum_{i=1}^k x_i = n$? ...