simplify (1+i/n)^3 — step-by-step solution | Explain MAths

simplify (1+i/n)^3

The original denominators must be nonzero:

n \in ℝ \ {0}

Let's reduce:

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

With Newton's binomial formula:

{(a + b)}^{n} = \sum_{k = 0}^{n} (\binom{n!}{k! (n - k)!}) a^{n - k} b^{k}

, let's expand the expression:

\frac{3!}{3! \cdot 0!} \cdot 1^{3} \cdot {(\frac{i}{n})}^{0}

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

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

+ \frac{3!}{0! \cdot 3!} \cdot 1^{0} \cdot {(\frac{i}{n})}^{3}

ℹ

{(\frac{i}{n})}^{0}

=

1

‖

1^{0}

=

1

\frac{3!}{3! \cdot 0!}

+ \frac{3!}{2! \cdot 1!} \cdot \frac{i}{n}

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

+ \frac{3!}{0! \cdot 3!} \cdot {(\frac{i}{n})}^{3}

ℹ

6 \cdot 1

=

6

‖

2 \cdot 1

=

2

‖

1 \cdot 2

=

2

‖

1 \cdot 6

=

6

\frac{6}{6}

+ \frac{6}{2} \cdot \frac{i}{n}

+ \frac{6}{2} \cdot {(\frac{i}{n})}^{2}

+ \frac{6}{6} \cdot {(\frac{i}{n})}^{3}

ℹ

\frac{6}{6}

=

1

‖

\frac{6}{2}

=

3

‖

3 \cdot \frac{i}{n}

=

\frac{3 i}{n}

‖

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

=

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

1

+ \frac{3 i}{n}

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

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

ℹ

{(\frac{i}{n})}^{2}

=

\frac{i^{2}}{n^{2}}

‖

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

=

\frac{i^{3}}{n^{3}}

1

+ \frac{3 i}{n}

+ 3 \cdot \frac{i^{2}}{n^{2}}

+ \frac{i^{3}}{n^{3}}

ℹ

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

=

\frac{3 i^{2}}{n^{2}}

‖

i^{2}

=

-1

‖

3 \cdot (-1)

=

-3

‖

i^{3}

=

- i

1

+ \frac{3 i}{n}

- \frac{3}{n^{2}}

- \frac{i}{n^{3}}

▶Simplified form:

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