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

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

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

expand

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

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

Solution steps

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{2}{2} = 1

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

\frac{( - 1 ) ^{n} n ( n - 1 ) x ^{n - 2}}{2 \cdot 5 ^{n + 1}}

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

(x + (\frac{1}{x})) \cdot (x - (\frac{1}{x}))

\frac{1}{2} (θ + \frac{sin ( 2 θ )}{2})

\frac{5}{( δ ^{2} + 3 ) ( s + 2 )}

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