What is A*(((1+i)^n-1)/(i*(1+i)^n)) ?

The solution to A*(((1+i)^n-1)/(i*(1+i)^n)) is (A((1+i)^n-1))/(i(1+i)^n)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify A(((1+i)^n-1)/(i(1+i)^n))

simplify

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

\frac{A ( ( 1 + i ) ^{n} - 1 )}{i ( 1 + i ) ^{n}}

Solution steps

A (\frac{( 1 + i ) ^{n} - 1}{i ( 1 + i ) ^{n}})

Apply rule:

(a) = a

(\frac{( 1 + i ) ^{n} - 1}{i ( 1 + i ) ^{n}}) = \frac{( 1 + i ) ^{n} - 1}{i ( 1 + i ) ^{n}}

= A \frac{( 1 + i ) ^{n} - 1}{i ( 1 + i ) ^{n}}

Apply the fraction rule:

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

= \frac{A ( ( 1 + i ) ^{n} - 1 )}{i ( 1 + i ) ^{n}}

(a - 1) (a + 2) - (a - 3) (a - 1)

\frac{cos ^{2} ( x )}{sin ( x )}

\frac{tan ^{2} ( x )}{sec ^{2} ( x )}

\frac{5 ^{3}}{5 ^{d}}

(9^{2} + 1) - 2^{4} \cdot 16 + (12 + 3 \cdot 5) - 7

What is A*(((1+i)^n-1)/(i*(1+i)^n)) ?
The solution to A*(((1+i)^n-1)/(i*(1+i)^n)) is (A((1+i)^n-1))/(i(1+i)^n)