Matematické Fórum

Pavel · 04. 10. 2010 11:58 — Editoval Pavel (09. 10. 2010 13:11)

Inspirován úlohou Binomická čísla II. uvádím další:

Nechť $n\in\mathbb N$ . Najděte součet

$\Large{4n\choose 0}-{4n\choose 4}+{4n\choose 8}-{4n\choose 12}+\dots+(-1)^n{4n\choose 4n-8}-(-1)^n{4n\choose 4n-4}+(-1)^n{4n\choose 4n}$ .

HINT :

Autor skryl část textu. Zobrazit/skrýt!

andrew · 29. 10. 2010 20:01 — Editoval andrew (29. 10. 2010 20:05)

Autor skryl část textu. Zobrazit/skrýt!

Pavel · 23. 12. 2010 19:22 — Editoval Pavel (23. 12. 2010 19:23)

↑ andrew:

Po drahné době se opět dostávám k tomuto problému.

Bohužel to není dobře. Neplatí totiž

$\Re\Big[(1+ \sqrt{\mathrm{i}})^{4n}\Big] = {4n \choose 0} - {4n \choose 4} + {4n \choose 8} - {4n \choose 12} + \cdots +(-1)^n {4n \choose 4n}$

Stačí vzít druhý člen binomického rozvoje $(1+ \sqrt{\mathrm{i}})^{4n}$

$\binom {4n}{1}\sqrt i=\binom {4n}{1}\left(\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right)$

a je zřejmé, že i v něm se objevuje nenulová reálná složka $\binom {4n}{1}\,\frac{\sqrt 2}{2}$ , která ve výše uvedené identitě chybí.

Pavel · 23. 12. 2010 22:18 — Editoval Pavel (23. 12. 2010 22:24)

Řešení:

Nechť

$\varepsilon=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}$

je kořenem rovnice $x^8=1$ . Pak všechny kořeny této rovnice je možné vyjádřit ve tvaru $1,\varepsilon,\varepsilon^2,\varepsilon^3,\varepsilon^4,\varepsilon^5,\varepsilon^6,\varepsilon^7$ .

Sečtěme binomické rozvoje

$(1+\varepsilon)^{4n}=\binom{4n}{0}\varepsilon^0+\binom{4n}{1}\varepsilon+\binom{4n}{2}\varepsilon^2+\binom{4n}{3}\varepsilon^3+\dots+\binom{4n}{4n-3}\varepsilon^{4n-3}+\binom{4n}{4n-2}\varepsilon^{4n-2}+\binom{4n}{4n-1}\varepsilon^{4n-1}+\binom{4n}{4n}\varepsilon^{4n}\nl (1-\varepsilon)^{4n}=\binom{4n}{0}\varepsilon^0-\binom{4n}{1}\varepsilon+\binom{4n}{2}\varepsilon^2-\binom{4n}{3}\varepsilon^3+\dots-\binom{4n}{4n-3}\varepsilon^{4n-3}+\binom{4n}{4n-2}\varepsilon^{4n-2}-\binom{4n}{4n-1}\varepsilon^{4n-1}+\binom{4n}{4n}\varepsilon^{4n}\nl \nl (1+\varepsilon)^{4n}+(1-\varepsilon)^{4n}=2\left[\binom{4n}{0}\varepsilon^0+\binom{4n}{2}\varepsilon^2+\binom{4n}{4}\varepsilon^4+\binom{4n}{6}\varepsilon^6+\dots+\binom{4n}{4n-6}\varepsilon^{4n-6}+\binom{4n}{4n-4}\varepsilon^{4n-4}+\binom{4n}{4n-2}\varepsilon^{4n-2}+\binom{4n}{4n}\varepsilon^{4n}\right]$

$(1+\varepsilon^3)^{4n}=\binom{4n}{0}\varepsilon^0+\binom{4n}{1}\varepsilon^3+\binom{4n}{2}\varepsilon^6+\binom{4n}{3}\varepsilon^9+\dots+\binom{4n}{4n-3}\varepsilon^{12n-9}+\binom{4n}{4n-2}\varepsilon^{12n-6}+\binom{4n}{4n-1}\varepsilon^{12n-3}+\binom{4n}{4n}\varepsilon^{12n}\nl (1-\varepsilon^3)^{4n}=\binom{4n}{0}\varepsilon^0-\binom{4n}{1}\varepsilon^3+\binom{4n}{2}\varepsilon^6-\binom{4n}{3}\varepsilon^9+\dots-\binom{4n}{4n-3}\varepsilon^{12n-9}+\binom{4n}{4n-2}\varepsilon^{12n-6}-\binom{4n}{4n-1}\varepsilon^{12n-3}+\binom{4n}{4n}\varepsilon^{12n}\nl \nl (1+\varepsilon^3)^{4n}+(1-\varepsilon^3)^{4n}=2\left[\binom{4n}{0}\varepsilon^0+\binom{4n}{2}\varepsilon^6+\binom{4n}{4}\varepsilon^{12}+\dots+\binom{4n}{4n-4}\varepsilon^{12n-12}+\binom{4n}{4n-2}\varepsilon^{12n-6}+\binom{4n}{4n}\varepsilon^{12n}\right]$

Oba součty sečtěme

$(1+\varepsilon)^{4n}+(1-\varepsilon)^{4n}+(1+\varepsilon^3)^{4n}+(1-\varepsilon^3)^{4n}=2\left[\binom{4n}{0}(\varepsilon^0+\varepsilon^0)+\binom{4n}{2}(\varepsilon^2+\varepsilon^6)+\binom{4n}{4}(\varepsilon^4+\varepsilon^{12})+\binom{4n}{6}(\varepsilon^6+\varepsilon^{18})+\dots+\right.\nl\left.+\binom{4n}{4n-6}(\varepsilon^{4n-6}+\varepsilon^{12n-18})+\binom{4n}{4n-4}(\varepsilon^{4n-4}+\varepsilon^{12n-12})+\binom{4n}{4n-2}(\varepsilon^{4n-2}+\varepsilon^{12n-6})+\binom{4n}{4n}(\varepsilon^{4n}+\varepsilon^{12n})\right]$

Součty mocnin kořene $\varepsilon$ lze zjednodušit podle následujícího pravidla:

$\varepsilon^{4n+2}+\varepsilon^{3(4n+2)}=\varepsilon^{4n+2}+\varepsilon^{12n+6}=\varepsilon^{4n+2}(1+\varepsilon^{8n+4})=\varepsilon^{4n+2}[1+(\varepsilon^4)^{2n+1}]=\varepsilon^{4n+2}[1+(-1)^{2n+1}]=0,\quad\forall n\in\mathbb{N},\nl \varepsilon^{4n}+\varepsilon^{3(4n)}=\varepsilon^{4n}+\varepsilon^{12n}=\varepsilon^{4n}(1+\varepsilon^{8n})=(\varepsilon^4)^n[1+(\varepsilon^8)^n]=(-1)^n(1+1^n)=2\cdot(-1)^n,\quad\forall n\in\mathbb{N}.$

Pak tedy platí

$(1+\varepsilon)^{4n}+(1-\varepsilon)^{4n}+(1+\varepsilon^3)^{4n}+(1-\varepsilon^3)^{4n}=\ldots=2\biggl[2\binom{4n}{0}-2\binom{4n}{4}+2\binom{4n}{8}-2\binom{4n}{12}+\dots+\nl +2\cdot(-1)^{n-3}\binom{4n}{4n-12}+2\cdot(-1)^{n-2}\binom{4n}{4n-8}+2\cdot(-1)^{n-1}\binom{4n}{4n-4}+2\cdot(-1)^n\binom{4n}{4n}\biggr]=\nl =4\left[\binom{4n}{0}-\binom{4n}{4}+\binom{4n}{8}-\binom{4n}{12}+\dots+(-1)^{n-3}\binom{4n}{4n-12}+(-1)^{n-2}\binom{4n}{4n-8}+(-1)^{n-1}\binom{4n}{4n-4}+(-1)^n\binom{4n}{4n}\right]$

Odtud určíme hledaný součet

$\binom{4n}{0}-\binom{4n}{4}+\binom{4n}{8}-\binom{4n}{12}+\dots+(-1)^{n-3}\binom{4n}{4n-12}+(-1)^{n-2}\binom{4n}{4n-8}+(-1)^{n-1}\binom{4n}{4n-4}+(-1)^n\binom{4n}{4n}=\nl=\frac 14\,\left[(1+\varepsilon)^{4n}+(1-\varepsilon)^{4n}+(1+\varepsilon^3)^{4n}+(1-\varepsilon^3)^{4n}\right]=\frac 14\left\{\left[\sqrt{2+\sqrt{2}}\,\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)\right]^{4n}+\left[\sqrt{2-\sqrt{2}}\,\left(\cos\left(-\frac{3\pi}{8}\right)+i\sin\left(-\frac{3\pi}{8}\right)\right)\right]^{4n}+\right.\nl\left.+\left[\sqrt{2-\sqrt{2}}\,\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}\right)\right]^{4n}+\left[\sqrt{2+\sqrt{2}}\,\left(\cos\left(-\frac{\pi}{8}\right)+i\sin\left(-\frac{\pi}{8}\right)\right)\right]^{4n}\right\}=\nl \frac 14\left[(2+\sqrt{2})^{2n}\,\left(\cos\frac{\pi n}{2}+i\sin\frac{\pi n}{2}\right)+(2-\sqrt{2})^{2n}\,\left(\cos\frac{3\pi n}{2}-i\sin\frac{3\pi n}{2}\right)+(2-\sqrt{2})^{2n}\,\left(\cos\frac{3\pi n}{2}+i\sin\frac{3\pi n}{2}\right)+(2+\sqrt{2})^{2n}\,\left(\cos\frac{\pi n}{2}-i\sin\frac{\pi n}{2}\right)\right]=\nl \frac 14\left[2(2+\sqrt{2})^{2n}\,\cos\frac{\pi n}{2}+2(2-\sqrt{2})^{2n}\,\cos\frac{3\pi n}{2}\right]=\frac 12\left[(2+\sqrt{2})^{2n}\,\cos\frac{\pi n}{2}+(2-\sqrt{2})^{2n}\,\cos\frac{3\pi n}{2}\right]=\nl=\frac 12\left[2^n(\sqrt{2}+1)^{2n}\,\cos\frac{\pi n}{2}+2^n(\sqrt{2}-1)^{2n}\,\cos\frac{3\pi n}{2}\right]=2^{n-1}\left[(\sqrt{2}+1)^{2n}\,\cos\frac{\pi n}{2}+(\sqrt{2}-1)^{2n}\,\cos\frac{3\pi n}{2}\right].$

Výsledek bude záviset na tom, zda je $n$ sudé nebo liché.

$\binom{4n}{0}-\binom{4n}{4}+\dots+(-1)^{n-1}\binom{4n}{4n-4}+(-1)^n\binom{4n}{4n}= \begin{cases} 0,\qquad n=2k+1,\nl (-1)^{\frac n2}\,2^{n-1}\left[(\sqrt{2}+1)^{2n}+(\sqrt{2}-1)^{2n}\right],\quad n=2k,\ k\in\mathbb{N}. \end{cases}$