Linked Questions

2
votes
1answer
224 views

closed form for $\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+…+\binom{n}{n}$ [duplicate]

closed form for $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$$ I tried to solve it by : $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}=\sum_{k=0}^{n/3}\binom{n}{3k}$$ $$\...
0
votes
0answers
57 views

Sum involving binomial $\sum_{k=0}^{n} \binom{3n}{3k}$ [duplicate]

The main question is to evaluate: $$\sum_{k=0}^{n} \binom{3n}{3k}$$ There is a standard technique but I cannot split the sums apart and then add them together. Could you help with this step?
53
votes
20answers
10k views

Interesting results easily achieved using complex numbers

I was just looking at a calculus textbook preparing my class for next week on complex numbers. I found it interesting to see as an exercise a way to calculate the usual freshman calculus integrals $\...
60
votes
14answers
7k views

Are there theoretical applications of trigonometry?

I am a high school student currently taking pre-calculus. We have just finished a unit on analytic trigonometry. Are any purely theoretical uses for trigonometry? More specifically, can ...
19
votes
8answers
14k views

Proving $\sum_{k=0}^n(-1)^k\binom nk=0$

Show that $$\sum_{k=0}^n(-1)^k\binom nk=0$$ So for odd $n$ we have an even number of terms. So $\binom nk=\binom n{n-k}$ which have opposite signs. Thus the sum is 0. For even $n$ we have that $$\...
13
votes
6answers
805 views

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 ...
29
votes
3answers
2k views

Combinatorial proof of a Stirling number identity.

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(-...
9
votes
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 ...
3
votes
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{...
5
votes
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 ...
6
votes
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
6
votes
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 ...
5
votes
1answer
2k views

Sum of every $k$th 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 ...
12
votes
1answer
890 views

Counting subsets with r mod 5 elements

Some time ago Qiaochu Yuan asked about counting subsets of a set whose number of elements is divisible by 3 (or 4). The story becomes even more interesting if one asks about number of subsets of n-...
1
vote
1answer
625 views

Counting descending sequences of positive integers

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 $\sum_{i=1}^k x_i = n$? ...

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