erweitern ((n+1)^4)/(2n^4)

Lösung

erweitern

\frac{( n + 1 ) ^{4}}{2 n ^{4}}

\frac{1}{2} + \frac{2}{n} + \frac{3}{n ^{2}} + \frac{2}{n ^{3}} + \frac{1}{2 n ^{4}}

Schritte zur Lösung

\frac{( n + 1 ) ^{4}}{2 n ^{4}}

(n + 1)^{4} = n^{4} + 4 n^{3} + 6 n^{2} + 4 n + 1

= \frac{n ^{4} + 4 n ^{3} + 6 n ^{2} + 4 n + 1}{2 n ^{4}}

Wende Bruchregel an:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{n ^{4} + 4 n ^{3} + 6 n ^{2} + 4 n + 1}{2 n ^{4}} = \frac{n ^{4}}{2 n ^{4}} + \frac{4 n ^{3}}{2 n ^{4}} + \frac{6 n ^{2}}{2 n ^{4}} + \frac{4 n}{2 n ^{4}} + \frac{1}{2 n ^{4}}

= \frac{n ^{4}}{2 n ^{4}} + \frac{4 n ^{3}}{2 n ^{4}} + \frac{6 n ^{2}}{2 n ^{4}} + \frac{4 n}{2 n ^{4}} + \frac{1}{2 n ^{4}}

\frac{n ^{4}}{2 n ^{4}} = \frac{1}{2}

\frac{4 n ^{3}}{2 n ^{4}} = \frac{2}{n}

\frac{6 n ^{2}}{2 n ^{4}} = \frac{3}{n ^{2}}

\frac{4 n}{2 n ^{4}} = \frac{2}{n ^{3}}

= \frac{1}{2} + \frac{2}{n} + \frac{3}{n ^{2}} + \frac{2}{n ^{3}} + \frac{1}{2 n ^{4}}

e x p an d (e^{- x} - 1)^{2}

e x p an d x (x + 4 + 2)

e x p an d \frac{π ( 3 ln ( 3 ) - 2 )}{2}

e x p an d c (n, n - 1)

e x p an d \frac{X}{( 1 - X ) ^{2}}