What is A*((1-(\frac{1+G)/(1+i))^n}{i-G}) ?

The solution to A*((1-(\frac{1+G)/(1+i))^n}{i-G}) is (A(2^n-((1+G)(1-i))^n)(-G-i))/(2^n(G^2+1))

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simplify A*((1-(\frac{1+G)/(1+i))^n}{i-G})

simplify

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

\frac{A ( 2 ^{n} - ( ( 1 + G ) ( 1 - i ) ) ^{n} ) ( - G - i )}{2 ^{n} ( G ^{2} + 1 )}

Solution steps

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

Apply rule:

(a) = a

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

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

Simplify

\frac{1 - ( \frac{1 + G}{1 + i} ) ^{n}}{i - G} : \frac{( 2 ^{n} - ( ( 1 + G ) ( 1 - i ) ) ^{n} ) ( - G - i )}{2 ^{n} ( G ^{2} + 1 )}

= A \frac{( 2 ^{n} - ( ( 1 + G ) ( 1 - i ) ) ^{n} ) ( - G - i )}{2 ^{n} ( G ^{2} + 1 )}

Apply the fraction rule:

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

= \frac{A ( 2 ^{n} - ( ( 1 + G ) ( 1 - i ) ) ^{n} ) ( - G - i )}{2 ^{n} ( G ^{2} + 1 )}

4 (2 a + 6) - 3 (4 a - 8)

\frac{( 2 - i ) ( 4 - 3 i ) ^{2}}{3 + 2 i}

\frac{4}{3 x ^{3}} + \frac{x}{6 x ^{3} + 3 x ^{2}}

(x^{3} - 2 x + 3) + (3 x^{2} + 6 x - 2)

\frac{8 x ^{3} - 27}{2 x - 3}

What is A*((1-(\frac{1+G)/(1+i))^n}{i-G}) ?
The solution to A*((1-(\frac{1+G)/(1+i))^n}{i-G}) is (A(2^n-((1+G)(1-i))^n)(-G-i))/(2^n(G^2+1))