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

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

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

expand

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

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

Solution steps

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

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

Divide the numbers:

\frac{2}{2} = 1

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

(1 + sin^{3} (t)) (3 sin^{2} (t) cos (t))

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

(x^{2} + 1) x

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

(1 + sin (x)) (1 - cos (x))

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