What is expand (2*3^k+3(-1)^{k-1}+3)/2 ?

The solution to expand (2*3^k+3(-1)^{k-1}+3)/2 is 3^k+(3(-1)^{k-1})/2+3/2

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expand (2*3^k+3(-1)^{k-1}+3)/2

expand

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

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

Solution steps

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{2}{2} = 1

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

(2) (3 s + 1) (s + \frac{1}{16})

\frac{x + 1}{( 2 x - 1 ) x ^{2}}

5 (2 x)

- \frac{g ( m t b - m g )}{b ( g + m )}

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

What is expand (2*3^k+3(-1)^{k-1}+3)/2 ?
The solution to expand (2*3^k+3(-1)^{k-1}+3)/2 is 3^k+(3(-1)^{k-1})/2+3/2