erweitern (1+(2i)/n)^3

Lösung

erweitern

(1 + \frac{2 i}{n})^{3}

\frac{n ^{2} - 12}{n ^{2}} + \frac{6 n ^{2} - 8}{n ^{3}} i

Schritte zur Lösung

(1 + \frac{2 i}{n})^{3}

Wende Formel für perfekte dritte Potenzen an:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

(1 + \frac{2 i}{n})^{3} = 1^{3} + 3 \cdot 1^{2} \cdot \frac{2 i}{n} + 3 \cdot 1 \cdot (\frac{2 i}{n})^{2} + (\frac{2 i}{n})^{3}

= 1^{3} + 3 \cdot 1^{2} \cdot \frac{2 i}{n} + 3 \cdot 1 \cdot (\frac{2 i}{n})^{2} + (\frac{2 i}{n})^{3}

1^{3} = 1

3 \cdot 1^{2} \cdot \frac{2 i}{n} = 3 \cdot \frac{2 i}{n}

3 \cdot 1 \cdot (\frac{2 i}{n})^{2} = - 3 \cdot \frac{4}{n ^{2}}

(\frac{2 i}{n})^{3} = - \frac{8 i}{n ^{3}}

= 1 + 3 \cdot \frac{2 i}{n} - 3 \cdot \frac{4}{n ^{2}} - \frac{8 i}{n ^{3}}

Vereinfache

1 + 3 \cdot \frac{2 i}{n} - 3 \cdot \frac{4}{n ^{2}} - \frac{8 i}{n ^{3}} : \frac{n ^{2} - 12}{n ^{2}} + \frac{6 n ^{2} - 8}{n ^{3}} i

= \frac{n ^{2} - 12}{n ^{2}} + \frac{6 n ^{2} - 8}{n ^{3}} i

s im pl i f y (8 + sin (t))^{2} + cos^{2} (t)

e x p an d sec (y) (tan (y) - sec (y)) d y

e x p an d H (8 - t) (sin (t - 1) - sin (t - 4))

e x p an d \frac{5 s ^{2} + 4}{( s ^{2} + 17 ) ( s - 1 )}

e x p an d e^{2 x} (x^{3} + 1)