簡約 (1+i/n)^3

解

簡約

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

\frac{n ^{2} - 3}{n ^{2}} + \frac{3 n ^{2} - 1}{n ^{3}} i

解答ステップ

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

完全立方式を適用する:

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

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

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

1^{3} = 1

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

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

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

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

簡素化

1 + 3 \cdot \frac{i}{n} - 3 \cdot \frac{1}{n ^{2}} - \frac{i}{n ^{3}} : \frac{n ^{2} - 3}{n ^{2}} + \frac{3 n ^{2} - 1}{n ^{3}} i

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