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my attempt:

let's put $2000 \choose 2000$+...+$2000 \choose 1001 $+...+$2000 \choose 8 $+$2000 \choose 5$+$2000 \choose 2$=$A+B$

with $A$=$2000 \choose 2000$+...+$2000 \choose 1001 $

and

$B$=$2000 \choose 998 $...+$2000 \choose 8 $+$2000 \choose 5$+$2000 \choose 2$

using this $ {n \choose k}= {n \choose n-k}$ we can make $A$=$2000 \choose 0$+$2000 \choose 3$+$2000 \choose 6$...+$2000 \choose 999$,and we have $B$=$2000 \choose 998 $...+$2000 \choose 8 $+$2000 \choose 5$+$2000 \choose 2$

so $A+B$=$2000 \choose 0$+$2000 \choose 2$+$2000 \choose 3$+$2000 \choose 5$+...+$2000 \choose 999$=$\sum_{k=0}^{2000} { 2000 \choose k}-\sum_{k=0}^{2000} { 2000 \choose 3k+1}=2^{2000}-\frac{2^{2000}+2}{3}=2.\frac{2^{2000}-1}{3}$

beacause $\sum_{k=0}^{2000} { 2000 \choose 3k+1}=\frac{2^{2000}+2}{3}$.

does my attempt is correct?

RobPratt
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    How did you find out $\binom{2000}{3k+1}$? – insipidintegrator Jul 11 '22 at 15:22
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    What is the pattern? Start with $2$ and add $3$ for each term? – 温泽海 Jul 11 '22 at 15:22
  • – insipidintegrator https://math.stackexchange.com/q/1352413/1069990 –  Jul 11 '22 at 15:31
  • the pattern is $3k+2$ –  Jul 11 '22 at 15:31
  • You have algebraic errors. For instance, you have this $A+B=\binom{2000}{0}+\binom{2000}{2}+\binom{2000}{3}+\dots+\binom{2000}{999}$ and try to say that this is equal to a summation $\sum\limits_{k=0}^{\color{red}{2000}}\binom{2000}{k}-\dots$ Your sum stopped at 999, not went all the way to 2000. Your final answer was essentially "taking every third entry of the sum $\binom{2000}{0}+\binom{2000}{1}+\binom{2000}{2}+\dots+\binom{2000}{2000}$ gives us just under two thirds the total of the sum" which of course sounds wrong. If we were to have guessed, we'd have guessed it be only one third – JMoravitz Jul 11 '22 at 15:39
  • You got the correct answer, which is $$[1, 0, 0]\begin{bmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 \end{bmatrix}^{2000} = [\dots, \dots, \frac 13 (2^{2000} - 1)]$$ This is question is equivalent to asking "how many subsets of $\dots 2000$ have size $3k+2$". – DanielV Jul 11 '22 at 15:52
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    @DanielV the OP has a multiplication of two there if you didn't notice. You say "you got the correct answer" and then cite an answer which is different than the OP's. – JMoravitz Jul 11 '22 at 16:05
  • DnieLV no it is not the same answer , iknow iam wrong but what i dont know is where is my mistake –  Jul 11 '22 at 16:12
  • @JMoravitz yeah now i understanf what you say, but can you edit my answer to be correct? –  Jul 11 '22 at 16:14
  • An explanation of what I said? What needs explanation? Do you recognize that $1+2+3+4+5+6$ is different than $1+2+3+4+5+6+7+8+9+10+11+12$? The first sum ended much earlier than the other sum. Do you recognize that $\binom{2000}{0}+\binom{2000}{2}+\binom{2000}{3}+\dots+\binom{2000}{999}$ ends much sooner than $\binom{2000}{0}+\binom{2000}{2}+\binom{2000}{3}+\dots+\binom{2000}{999}\color{red}{+\binom{2000}{1000}+\binom{2000}{1002}+\dots+\binom{2000}{2000}}$? This has all of these red numbers as well – JMoravitz Jul 11 '22 at 16:14
  • yeah i said you are right –  Jul 11 '22 at 16:15
  • As for how to fix... recognize that$\binom{2000}{2}+\binom{2000}{5}+\dots+\binom{2000}{2000}$ is by using $\binom{n}{k}=\binom{n}{n-k}$ just $\binom{2000}{1998}+\binom{2000}{1995}+\dots+\binom{2000}{3}+\binom{2000}{0}$, so if we called $X$ the original sum we have $2X$ is equal to the sum of everything but the $3k+1$ terms. Continue like you were trying to do, this time being allowed to use the same limits on the summation correctly, and get the value of $2X$ at the end, completing the problem to get the value of $X$ by dividing that result by two. – JMoravitz Jul 11 '22 at 16:22
  • @JMoravitz You are right, I missed "2." – DanielV Jul 11 '22 at 17:54

3 Answers3

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$$S={2000 \choose 2} + {2000 \choose 5} + {2000 \choose 8} + {2000 \choose 11} + \cdots + {2000 \choose 2000}$$

Consider the function $f(x)=x(1+x)^{2000}$. Applying roots of unity filter,

$$S=\frac{2^{2000}+\omega(1+w)^{2000}+\omega^2(1+w^2)^{2000}}3=\frac{2^{2000}+\omega^2+\omega}3=\frac{2^{2000}-1}3$$

Note that a factor of $x$ has been multiplied with our function in question to account for the offset of the starting point.

DatBoi
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  • PS: Other users have pointed out the errors in OP's solution. My answer is purely for the sake of providing a simple solution to the question, which may be helpful for future readers. – DatBoi Jul 11 '22 at 18:54
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Let $C=\sum_{k=0}^{666}{2000 \choose 3k+2}$, using symmetry of binomial coefficients $C=\sum_{k=0}^{666}{2000 \choose 3k}$. Let $B=\sum_{k=0}^{666}{2000 \choose 3k+1}$.

$$2^{2000}=\sum_{n=0}^{2000}{2000 \choose n}=\\\sum_{k=0}^{666}{2000 \choose 3k}+\sum_{k=0}^{666}{2000 \choose 3k+1}+\sum_{k=0}^{666}{2000 \choose 3k+2}=2C+B.$$

$$\left(\frac12+i\frac{\sqrt3}2\right)^{2000}=\left(1+\left(-\frac12+i\frac{\sqrt3}2\right)\right)^{2000}= \sum_{n=0}^{2000}{2000 \choose n}\left(-\frac12+i\frac{\sqrt3}2\right)^n.$$

$$\left(-\frac12+i\frac{\sqrt3}2\right)^3=1\Rightarrow\\ \sum_{n=0}^{2000}{2000 \choose n}\left(-\frac12+i\frac{\sqrt3}2\right)^n=C+B\left(-\frac12+i\frac{\sqrt3}2\right)+C\left(-\frac12+i\frac{\sqrt3}2\right)^2=\\ B\left(-\frac12+i\frac{\sqrt3}2\right)+C\left(\frac12-i\frac{\sqrt3}2\right)=(B-C)\left(-\frac12+i\frac{\sqrt3}2\right).$$

$$\left(\frac12+i\frac{\sqrt3}2\right)^{6}=1\Rightarrow \left(\frac12+i\frac{\sqrt3}2\right)^{2000}=\left(\frac12+i\frac{\sqrt3}2\right)^{2}=-\frac{1}{2}+\frac{\sqrt3}2i.$$

$$(B-C)\left(-\frac12+i\frac{\sqrt3}2\right)=-\frac{1}{2}+\frac{\sqrt3}2i \Rightarrow B-C=1 \Rightarrow B=C+1.$$

$$2^{2000}=2C+B=3C+1 \Rightarrow C=\frac{2^{2000}-1}3.$$

Ivan Kaznacheyeu
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0

$${2000 \choose 0} + {2000 \choose 2} + {2000 \choose 3} + {2000 \choose 5} + \cdots + {2000 \choose 999}$$ $$ = \sum_{k=0}^{2000} {2000 \choose k} - \sum_{0 \le k <= 2000 \\ k\equiv 1 \pmod 3} {2000 \choose k}$$

This step is wrong.

DanielV
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