What is expand (2(-1)^k+3(-1)^{k-1}+2*3^{1+k}+5)/2 ?

The solution to expand (2(-1)^k+3(-1)^{k-1}+2*3^{1+k}+5)/2 is (-1)^k+(3(-1)^{k-1})/2+3^{1+k}+5/2

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expand (2(-1)^k+3(-1)^{k-1}+2*3^{1+k}+5)/2

expand

\frac{2 ( - 1 ) ^{k} + 3 ( - 1 ) ^{k - 1} + 2 \cdot 3 ^{1 + k} + 5}{2}

(- 1)^{k} + \frac{3 ( - 1 ) ^{k - 1}}{2} + 3^{1 + k} + \frac{5}{2}

Solution steps

\frac{2 ( - 1 ) ^{k} + 3 ( - 1 ) ^{k - 1} + 2 \cdot 3 ^{1 + k} + 5}{2}

Apply the fraction rule:

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

\frac{2 ( - 1 ) ^{k} + 3 ( - 1 ) ^{k - 1} + 2 \cdot 3 ^{1 + k} + 5}{2} = \frac{2 ( - 1 ) ^{k}}{2} + \frac{3 ( - 1 ) ^{k - 1}}{2} + \frac{2 \cdot 3 ^{1 + k}}{2} + \frac{5}{2}

= \frac{2 ( - 1 ) ^{k}}{2} + \frac{3 ( - 1 ) ^{k - 1}}{2} + \frac{2 \cdot 3 ^{k + 1}}{2} + \frac{5}{2}

Divide the numbers:

\frac{2}{2} = 1

= (- 1)^{k} + \frac{3 ( - 1 ) ^{k - 1}}{2} + 3^{k + 1} + \frac{5}{2}

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What is expand (2(-1)^k+3(-1)^{k-1}+2*3^{1+k}+5)/2 ?
The solution to expand (2(-1)^k+3(-1)^{k-1}+2*3^{1+k}+5)/2 is (-1)^k+(3(-1)^{k-1})/2+3^{1+k}+5/2