5

I am trying to find $$\binom{1000}{3}+\binom{1000}{8}+\binom{1000}{13}+\dots+\binom{1000}{998}=?$$

My work:

Let $\omega=\exp(\displaystyle\frac{2\pi i}{5}) $ and so $\omega^5=1$ . Then $1 + \omega+\omega^2+\omega^3+\omega^4=0$

$$(1+1)^{1000}+w^2(1+w)^{1000}+w^4(1+w^2)^{1000}+w^6(1+w^3)^{1000}+w^8(1+w^4)^{1000}\\=\sum_{k=0}^{1000}\bigg[\binom{1000}{k}+\binom{1000}{k}w^{k+2}+\binom{1000}{k}w^{2k+4}+\binom{1000}{k}w^{3k+6}+\binom{1000}{k}w^{4k+8}\bigg]\\= 5\bigg[\binom{1000}{3}+\binom{1000}{8}+\binom{1000}{13}+\dots+\binom{1000}{998}\bigg]$$

However , when I come to calculate the result of $(1+1)^{1000}+w^2(1+w)^{1000}+w^4(1+w^2)^{1000}+w^6(1+w^3)^{1000}+w^8(1+w^4)^{1000}$ , I stuck in it , because I could not simplify it using $\omega^5=1$ or $1 + \omega+\omega^2+\omega^3+\omega^4=0$.

Hence , I am looking for helps to find a closed formula for the binomial expansion by simplifying $(1+1)^{1000}+w^2(1+w)^{1000}+w^4(1+w^2)^{1000}+w^6(1+w^3)^{1000}+w^8(1+w^4)^{1000}$

ADDENTUM: I want to reach an integer solution, as it is expected from this expression. For example, Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$ , answer of this question is $(2^{2000}+2)/3$. It is what kind of answer I want to reach. So, can you help me to simplify given expression into this type of integer result answer ?

Thanks in advance !!

hbghlyj
  • 2,115
  • @insipidintegrator no it does not answer my question .. please try to see my problem –  Jul 12 '22 at 19:28
  • @insipidintegrator my problem is on simplification , before wrting a comment , try to read the question –  Jul 12 '22 at 19:30
  • Do you see how to show $(1+w)^{1000}=(w^{-2}+w^2)^{1000}$? – Brian Moehring Jul 12 '22 at 19:48
  • @BrianMoehring i cannot see it directly , how did you derive it ? –  Jul 12 '22 at 19:50
  • 2
    In short, multiply the left side by $(w^2)^{1000}=1$ and then note $w^3=w^{-2}$. Do you see why that's more helpful as a form, at least to get an expression in terms of real numbers? – Brian Moehring Jul 12 '22 at 20:01
  • @BrianMoehring i obtained $$w^2(1+w)^{1000}+w^4(1+w^2)^{1000}+w(1+w^3)^{1000}+w^3(1+w^4)^{1000}$$ However , this caused another problem to simplify –  Jul 12 '22 at 20:28
  • In general, there is no reason to expect you would be able to simplify to that kind of answer as in the other question. Cube roots, as in the other question, may be a special case. – Ted Jul 12 '22 at 22:02
  • In the case of cube roots, $\cos(2\pi/3)=-1/2$ is rational, which is why we can give that answer in a nicer form. You can write your answer in terms of $\sqrt{5}$ and rationals, but that's as simple a closed form as one can expect. When I said you'd need to specify if you needed an efficient solution in terms of integer arithmetic, I was meaning something like fast matrix exponentiation to efficiently evaluate a linear recursion (which might be necessary if you're trying to compute the answer mod $n$ for some $n$) – Brian Moehring Jul 12 '22 at 22:18

3 Answers3

2

For comfort of notation (I am more inclined to make a mistake using $\omega$ because I am accustomed with it being a third root of unity), let $\displaystyle \alpha=e^{\frac{2\pi i}{5}}$.

Then, $\alpha^5=1$ and $\alpha^4+ \alpha^3+ \alpha^2+ \alpha+1=0$. Dividing the latter equation by $\alpha^2$ gives $$\left(\alpha^2+\frac{1}{\alpha^2}\right)+\left(\alpha+\frac{1}{\alpha}\right)+1=0$$ Let $\displaystyle{\alpha}+\frac{1}{\alpha}=u$. Then $$u^2-2+u+1=0\implies u^2+u-1=0$$ and thus $$\displaystyle u=\frac{-1\pm \sqrt 5}{2}$$

Now, the original expression is $$2^{1000}+ {\alpha}^2(1+{\alpha})^{1000}+ \alpha ^4(1+ \alpha ^2)^{1000}+ \alpha ^6(1+ \alpha^3 )^{1000}+ \alpha ^8(1+ \alpha^4 )^{1000}$$$$= 2^{1000}+ {\alpha}^2(1+{\alpha})^{1000}+ \frac{1}{\alpha}( 1+ \alpha ^2)^{1000}+ \alpha(1+ \alpha^3 )^{1000}+ \frac{1}{\alpha^2}(1+ \alpha^4 )^{1000}$$, all using $\alpha^5=1$. Now, put $\alpha^3=\frac{1}{\alpha^2}$ and $\alpha^4=\frac{1}{\alpha}$ so you get $$= 2^{1000}+ {\alpha}^2(1+{\alpha})^{1000}+ \frac{1}{\alpha}( 1+ \alpha ^2)^{1000}+ \alpha(1+ \alpha^2)^{1000}+ \frac{1}{\alpha^2}(1+ \alpha)^{1000}$$ using $\alpha^{1000}=\alpha^{2000}=1$. Now write this as $$2^{1000}+(u^2-2)(1+\alpha)^{1000}+u(1+\alpha^2)^{1000}$$ Divide the last term by $\alpha^{1000}$:$$2^{1000}+(u^2-2)(1+\alpha)^{1000}+u^{1001}$$ Now, let $\alpha=\cos\theta+i\sin\theta$ where $5\theta=2\pi$. Thus, $$(1+\alpha)^{1000}=(1+\cos\theta+i\sin\theta)^{1000}$$$$=(2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2})^{1000}=2^{1000}\left(\cos\frac{\theta}{2}\right)^{1000}\left(\cos\frac{\theta}{2}+2i\sin\frac{\theta}{2}\right)^{1000}= 2^{1000}\left(\cos\frac{\theta}{2}\right)^{1000}\left(e^{\frac{2\pi i}{5}}\right)^{1000}= 2^{1000}\left(\cos\frac{\theta}{2}\right)^{1000}. $$

Finally, assimilating everything, we have

Expression $ = 2^{1000}+(u^2-2) 2^{1000}\left(\cos\frac{\pi}{5}\right)^{1000}+u^{1001}. $

Now, to decide the value of $u$, notice that $\alpha+\frac{1}{\alpha}=2\cos\frac{2\pi}{5}=\frac{\sqrt5-1}{2}$.

Ketan
  • 99
insipidintegrator
  • 6,642
  • It seems that this answer gives non-integer value because of cosine and "u" ,but the expected answer must be integer. How will you solve this issue ? –  Jul 12 '22 at 20:31
  • 3
    @ulahh As a mathematical problem, there is no issue to solve. E.g. if you've ever seen Binet's formula for Fibonacci numbers, you've seen another instance when the closed form of an integer includes powers of non-integers. If your context requires integer arithmetic, there may be a way of efficiently calculating your value (I haven't tried) but that requirement would definitely need to be in the original post. – Brian Moehring Jul 12 '22 at 20:54
  • @BrianMoehring i have edited –  Jul 12 '22 at 21:33
  • To get closer to an integer here, note that $u=\phi-1$, and $u^2=\phi^2-2\phi+1=2-\phi$. Also, $\cos \frac{\tau}{10} = \frac{\phi}{2}$. A bit of rearrangement gives:

    $$N=2^{1000}-\phi^{1001}+\left(\frac{1}{\phi}\right)^{1000}$$

    Now the fun part: $\phi^{k+1}-\left(\frac{1}{\phi}\right)^k \approx L_k$, that is, the $k$th Lucas number. The final solution then is $2^{1000}-L_{1001}$.

     – Eric Snyder Jul 13 '22 at 04:25
  • Apologies; I missed a factor of $5$ in there somehow. The solution is $\frac15 (2^{1000} - L_{1001})$ – Eric Snyder Jul 13 '22 at 23:36
1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{{1000 \choose 3} + {1000 \choose 8} + {1000 \choose 13} + \cdots + {1000 \choose 998}} \\[5mm] = & \ \sum_{n = 0}^{199}{1000 \choose 5n + 3} = \sum_{n = 0}^{\infty}{1000 \choose 997 - 5n} \\[5mm] = & \ \sum_{n = 0}^{\infty}\ \oint_{\verts{z}\ =\ 1^{-}} {\pars{1 + z}^{1000} \over z^{998 - 5n}} {\dd z \over 2\pi\ic} \\[5mm] = & \ \oint_{\verts{z}\ =\ 1^{-}} {\pars{1 + z}^{1000} \over z^{998}} \sum_{n = 0}^{\infty}\pars{z^{5}}^{n} \,{\dd z \over 2\pi\ic} \\[5mm] = & \ \oint_{\verts{z}\ =\ 1^{-}} {\pars{1 + z}^{1000} \over z^{998}} {1 \over 1 - z^{5}}\,{\dd z \over 2\pi\ic} \\[5mm] \stackrel{z\ \mapsto\ 1/z}{=}\,\, & \ \oint_{\verts{z}\ =\ 1^{+}} {z\pars{1 + z}^{1000} \over z^{5} - 1} {\dd z \over 2\pi\ic} \\[5mm] = & \ \left.{1 \over 5}\sum_{n = -2}^{2}\xi_{n}^{2}\ \pars{1 + \xi_{n}}^{1000}\,\,\right\vert _{\,\xi_{n}\ \equiv\ \exp\pars{2n\pi\ic/5}} \\[5mm] = & \ \bbx{\color{#44f}{\begin{array}{l} \ds{{2^{1000} \over 5} + {2 \over 5}\,\Re\bracks{\expo{2\pi\ic/5}\pars{1 + \expo{2\pi\ic/5}}^{1000}\,}} \\[2mm] \ds{+\ {2 \over 5}\,\Re\bracks{\expo{4\pi\ic/5} \pars{1 + \expo{4\pi\ic/5}}^{1000}\,}} \end{array}}} \\ & \end{align} The final result is, indeed, a big number $\ds{\approx 2.1430 \times 10^{300}}$.

Felix Marin
  • 89,464
0

Hint: observe that

$$(1+w^k)^5 = 1 + \binom{5}{1}w^k + \binom{5}2w^{2k}+\binom{5}{3}w^{3k}+\binom{5}{4}w^{4k}+1 $$ When $k=1,4$ we get $$1 + \binom{5}{1}w + \binom{5}2w^{2}+\binom{5}{3}w^{3}+\binom{5}{4}w^{4}+1$$ Because $\binom{5}{i}=\binom{5}{5-i}$

When $k=2,3$ we get $$1 + \binom{5}{1}w^2 + \binom{5}2w^{2}+\binom{5}{3}w^{1}+\binom{5}{4}w^{3}+1$$

You can write $(1+w^k)^{1000}= \left[(1+w^k)^5\right]^{200}$ And use the above to reduce your expression.

IrbidMath
  • 3,187
  • 12
  • 27