matematikk.net

Finn en eksplisitt formel for

[tex]S_N=\sum_{n=1}^N n^4[/tex]

[tex]E = mc^2[/tex]

espen180 skrev:Finn en eksplisitt formel for
[tex]S_N=\sum_{n=1}^N n^4[/tex]

skriver dette som (1+n)^5, der

[tex](1+n)^5=1+5n+10n^2+10n^3+5n^4+n^5[/tex]

da er

[tex]1^5+2^5+3^5+...+(1+n)^5=(n+1)\,+\,5(1+2+3+...+n)\,+\,10(1^2+2^2+3^2+...n^2)\,+\,10(1^3+2^3+3^3+...+n^3)\,+\,5(1^4+2^4+3^4+...+n^4)\,+\,1^5+2^5+3^5+...+n^5[/tex]

rydder opp:

[tex]5(1^4+2^4+3^4+...+n^4)\,=\,(n+1)-n-1\,-\,5\left(\frac{n(n+1)}{2}\right)\,-\,10\left(\frac{n(n+1)(2n+1)}{6}\right)\,-\,10\left(\frac{n^2(n+1)^2}{4}\right)[/tex]

[tex]\sum_{n=1}^N n^4=1^4+2^4+3^4+...+n^4\,=\frac{n(n+1)(2n+1)}{6}\cdot\left(\frac{3n^2+3n-1}{5}\right)[/tex]

[tex]\frac{{n\left( {2n + 1} \right)\left( {n + 1} \right)\left( {3{n^2} + 3n - 1} \right)}}{{30}}[/tex]

Eller

[tex]\frac{1}{5}{n^5} + \frac{1}{2}{n^4} + \frac{1}{3}{n^3} - \frac{1}{{30}}[/tex]

Ser bra ut.