What is ((1-(1/3)(1-(-1/2)^{n-1})))/2 ?

The solution to ((1-(1/3)(1-(-1/2)^{n-1})))/2 is (2+(-1/2)^{n-1})/6

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simplify ((1-(1/3)(1-(-1/2)^{n-1})))/2

simplify

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

\frac{2 + ( - \frac{1}{2} ) ^{n - 1}}{6}

Solution steps

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

Apply rule:

(a) = a

(\frac{1}{3}) = \frac{1}{3}

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

Simplify

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

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

Apply the fraction rule:

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

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

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 + ( - \frac{1}{2} ) ^{n - 1}}{6}

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What is ((1-(1/3)(1-(-1/2)^{n-1})))/2 ?
The solution to ((1-(1/3)(1-(-1/2)^{n-1})))/2 is (2+(-1/2)^{n-1})/6