matematikk.net

En morsom oppgave, tok meg litt tid før jeg fant en smart løsning.

Finn den eksakte summen av rekken:

[tex]\frac{1}{1\cdot 2\cdot 3\cdot 4}+\frac{1}{5\cdot 6\cdot7 \cdot 8}+...[/tex]

Jeg aner egentlig ikke om jeg bryter noen regler her og løsningen er sikkert veldig stygg, men ble veldig oppslukt av denne oppgaven. er det feil så griner jeg

$\frac{1}{1*2*3*4} + \frac{1}{5*6*7*8} + \frac{1}{9*10*11*12} + \text{ ... } = \frac{1}{4!} + \frac{1}{\frac{8!}{4!}} + \frac{1}{\frac{12!}{8!}} + \text{ ... } = \frac{0!}{4!} + \frac{4!}{8!} + \frac{8!}{12!} + \text { ...}$

$a_n = \frac{\left( 4(n-1) \right)!}{(4n)!} = \frac{(4n - 4)!}{(4n)(4n - 1)(4n - 2)(4n - 3)(4n-4)!} = \frac{1}{4n(4n-1)(4n-2)(4n-3)}$


$\sum_{n = 1}^{\infty} \frac{1}{4n(4n-1)(4n-2)(4n-3)} = \sum_{k = 0}^{\infty}\frac{1}{(4k+4)(4k+3)(4k+2)(4k+1)}$

$= -\frac{1}{6}\sum_{k = 0}^{\infty}\frac{1}{4k+4} + \frac{1}{2}\sum_{k = 0}^{\infty}\frac{1}{4k + 3} - \frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{4k+2} + \frac{1}{6}\sum_{k = 0}^{\infty}\frac{1}{4k+1}$

$= -\frac{1}{6}\left(\sum_{k = 0}^{1}\frac{1}{4k+4} + \sum_{k = 1}^{\infty}\frac{1}{4k+4} \right) + \frac{1}{2}\left(\sum_{k = 0}^{1}\frac{1}{4k+3} + \sum_{k = 1}^{\infty}\frac{1}{4k+3} \right) -\frac{1}{2}\left(\sum_{k = 0}^{1}\frac{1}{4k+2} + \sum_{k = 1}^{\infty}\frac{1}{4k+2} \right) + \frac{1}{6}\left(\sum_{k = 0}^{1}\frac{1}{4k+1} + \sum_{k = 1}^{\infty}\frac{1}{4k+1} \right)$

ser leddene som går mot uendelig blir 0 så da står det igjen

$= -\frac{1}{6}\sum_{k = 0}^{1}\frac{1}{4k+4} + \frac{1}{2}\sum_{k = 0}^{1}\frac{1}{4k+3} -\frac{1}{2}\sum_{k = 0}^{1}\frac{1}{4k+2} + \frac{1}{6}\sum_{k = 0}^{1}\frac{1}{4k+1}$

$= -\frac{1}{6}\left(\frac{1}{4} + \frac{1}{8} \right) + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{7} \right) - \frac{1}{2}\left(\frac{1}{2} + \frac{1}{6}\right) + \frac{1}{6}\left( 1 + \frac{1}{5}\right)$

$= -\frac{1}{16} + \frac{5}{21} - \frac{1}{3} + \frac{1}{5} = \frac{71}{1680} \approx 0.0422619047619$

hallapaadeg skrev:Jeg aner egentlig ikke om jeg bryter noen regler her og løsningen er sikkert veldig stygg, men ble veldig oppslukt av denne oppgaven. er det feil så griner jeg

$\frac{1}{1*2*3*4} + \frac{1}{5*6*7*8} + \frac{1}{9*10*11*12} + \text{ ... } = \frac{1}{4!} + \frac{1}{\frac{8!}{4!}} + \frac{1}{\frac{12!}{8!}} + \text{ ... } = \frac{0!}{4!} + \frac{4!}{8!} + \frac{8!}{12!} + \text { ...}$

$a_n = \frac{\left( 4(n-1) \right)!}{(4n)!} = \frac{(4n - 4)!}{(4n)(4n - 1)(4n - 2)(4n - 3)(4n-4)!} = \frac{1}{4n(4n-1)(4n-2)(4n-3)}$


$\sum_{n = 1}^{\infty} \frac{1}{4n(4n-1)(4n-2)(4n-3)} = \sum_{k = 0}^{\infty}\frac{1}{(4k+4)(4k+3)(4k+2)(4k+1)}$

$= -\frac{1}{6}\sum_{k = 0}^{\infty}\frac{1}{4k+4} + \frac{1}{2}\sum_{k = 0}^{\infty}\frac{1}{4k + 3} - \frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{4k+2} + \frac{1}{6}\sum_{k = 0}^{\infty}\frac{1}{4k+1}$

$= -\frac{1}{6}\left(\sum_{k = 0}^{1}\frac{1}{4k+4} + \sum_{k = 1}^{\infty}\frac{1}{4k+4} \right) + \frac{1}{2}\left(\sum_{k = 0}^{1}\frac{1}{4k+3} + \sum_{k = 1}^{\infty}\frac{1}{4k+3} \right) -\frac{1}{2}\left(\sum_{k = 0}^{1}\frac{1}{4k+2} + \sum_{k = 1}^{\infty}\frac{1}{4k+2} \right) + \frac{1}{6}\left(\sum_{k = 0}^{1}\frac{1}{4k+1} + \sum_{k = 1}^{\infty}\frac{1}{4k+1} \right)$

ser leddene som går mot uendelig blir 0 så da står det igjen

$= -\frac{1}{6}\sum_{k = 0}^{1}\frac{1}{4k+4} + \frac{1}{2}\sum_{k = 0}^{1}\frac{1}{4k+3} -\frac{1}{2}\sum_{k = 0}^{1}\frac{1}{4k+2} + \frac{1}{6}\sum_{k = 0}^{1}\frac{1}{4k+1}$

$= -\frac{1}{6}\left(\frac{1}{4} + \frac{1}{8} \right) + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{7} \right) - \frac{1}{2}\left(\frac{1}{2} + \frac{1}{6}\right) + \frac{1}{6}\left( 1 + \frac{1}{5}\right)$

$= -\frac{1}{16} + \frac{5}{21} - \frac{1}{3} + \frac{1}{5} = \frac{71}{1680} \approx 0.0422619047619$

Såfremt jeg ikke er helt ute og kjøre, ser dette riktig ut

Edit: Wolfram får også [tex]\frac{71}{1680}[/tex].

Her er min løsning:
[tex]\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=\int_{0}^{1}\int_{0}^{x}\int_{0}^{x_1}\int_{0}^{x_2}=x_{3}^{4n}\, dx_3dx_2dx_1dx=\frac{1}{6}\int_{0}^{1}(1-x)^3x^{4n}dx.[/tex]

Derav [tex]S=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^3}{1-x^4}dx=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^2}{(x+1)(x^2+1)}dx[/tex]

Som kan dekompneres til: [tex]\frac{1}{6}\int_{0}^{1}(\frac{2}{x+1}-\frac{x+1}{x^2+1})dy.[/tex]

Ved å løse denne får vi:
[tex]\frac{1}{6}\left [ 2\ln(x+1)-\frac{1}{2}\ln(x^2+1)-\tan^{-1}x \right ]_{0}^{1}\\\\=\frac{1}{6}\left [ 2\ln2-\frac{1}{2}\ln2-\frac{\pi}{4} \right ]=\frac{1}{6}(\frac{3}{2}\ln2 -\frac{\pi}{4})=\frac{1}{4}\ln2-\frac{\pi}{24}\approx 0.042[/tex]