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${2000\choose1}+{2000\choose4}+{2000\choose7}+\cdots +{2000\choose1996}+{2000\choose1999}=?$

N. F. Taussig
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NHeeSeok
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    Welcome to MathSE. When you pose a question here, it is expeced that you include any work you have done on the problem and indicate where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Jul 07 '15 at 09:42
  • Is this a definition? Have you called '?' to this number? – ajotatxe Jul 07 '15 at 09:49
  • See this related question. There the upstairs value is constrained to be divisible by three, but the technique is the same. This is another older question on our site. You need to use different roots of unity as multipliers (all explained by discrete Fourier transforms). – Jyrki Lahtonen Jul 07 '15 at 10:31

2 Answers2

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$$(1+x)^n=\sum_{r=0}^n\binom nr x^r$$ where $3\nmid n$

Set $x=1,w,w^2$ where $w$ is a complex cube root of unity

$$P=(1+1)^n=\sum_{r=0}^n\binom nr 1^r$$

$$Q=(1+w)^n=\sum_{r=0}^n\binom nr w^r$$

$$R=(1+w^2)^n=\sum_{r=0}^n\binom nr w^{2r}$$

Find $P+w^2\cdot Q+w\cdot R$

Use $1+w+w^2=0$ to simplify the calculation

lab bhattacharjee
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a = ${2000\choose1}+{2000\choose4}+{2000\choose7}+\cdots +{2000\choose1996}+{2000\choose1999}$

b = ${2000\choose0}+{2000\choose3}+{2000\choose6}+\cdots +{2000\choose1995}+{2000\choose1998} = {2000\choose2}+{2000\choose5}+{2000\choose8}+\cdots +{2000\choose1997}+{2000\choose2000}$

and that $b = \dfrac{(1+1)^{2000} + (1+\omega)^{2000} + (1+\omega^2)^{2000}}{3}$. where $\omega$ is the primitive third root of unity. It can be seen that $1+\omega$ is the primitive sixth root of unity and $(1+\omega^2)$ is its inverse

So $(1+\omega)^{2000} = (1+\omega)^2 = \omega$

Similarly $(1+\omega^2)^{2000} = \omega^2$

Therefore, $b = \frac{2^{2000}-1}{3}$

$a+2b = 2^{2000}$

$a = 2^{2000} - 2b$

$a = 2^{2000} - 2\frac{2^{2000}-1}{3} = \frac{2^{2000}+2}{3} = \frac{2^{2000}+2}{3}$

Satish Ramanathan
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