What is ((m^{k+1}-1)/2)((m^{k+1}+1)/4) ?

The solution to ((m^{k+1}-1)/2)((m^{k+1}+1)/4) is ((m^{k+1}-1)(m^{k+1}+1))/8

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simplify ((m^{k+1}-1)/2)((m^{k+1}+1)/4)

simplify

(\frac{m ^{k + 1} - 1}{2}) (\frac{m ^{k + 1} + 1}{4})

\frac{( m ^{k + 1} - 1 ) ( m ^{k + 1} + 1 )}{8}

Solution steps

(\frac{m ^{k + 1} - 1}{2}) (\frac{m ^{k + 1} + 1}{4})

Simplify

(\frac{m ^{k + 1} - 1}{2}) (\frac{m ^{k + 1} + 1}{4}) : \frac{m ^{k + 1} - 1}{2} \cdot \frac{m ^{k + 1} + 1}{4}

= \frac{m ^{k + 1} - 1}{2} \cdot \frac{m ^{k + 1} + 1}{4}

Apply the fraction rule:

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

= \frac{( m ^{k + 1} - 1 ) ( m ^{k + 1} + 1 )}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{( m ^{k + 1} - 1 ) ( m ^{k + 1} + 1 )}{8}

f (\frac{1}{x})

2 x^{7} \cdot y - 54 x^{4} \cdot y^{4}

\frac{( 4 ^{x} \cdot 1 6 ^{x + 1} )}{( 4 ^{x - 1} \cdot 1 6 ^{x - 1} )}

3 q (3 q^{1}) (3 q - 1)

x^{3} + y^{3} + 2 x + 2 y

What is ((m^{k+1}-1)/2)((m^{k+1}+1)/4) ?
The solution to ((m^{k+1}-1)/2)((m^{k+1}+1)/4) is ((m^{k+1}-1)(m^{k+1}+1))/8