What is expand (7*2^{(k+1)}+(-1)^{(k+2)})/3 ?

The solution to expand (7*2^{(k+1)}+(-1)^{(k+2)})/3 is (7*2^{k+1})/3+((-1)^{k+2})/3

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expand (7*2^{(k+1)}+(-1)^{(k+2)})/3

expand

\frac{7 \cdot 2 ^{(k + 1)} + ( - 1 ) ^{(k + 2)}}{3}

\frac{7 \cdot 2 ^{k + 1}}{3} + \frac{( - 1 ) ^{k + 2}}{3}

Solution steps

\frac{7 \cdot 2 ^{k + 1} + ( - 1 ) ^{k + 2}}{3}

Apply the fraction rule:

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

\frac{7 \cdot 2 ^{k + 1} + ( - 1 ) ^{k + 2}}{3} = \frac{7 \cdot 2 ^{k + 1}}{3} + \frac{( - 1 ) ^{k + 2}}{3}

= \frac{7 \cdot 2 ^{k + 1}}{3} + \frac{( - 1 ) ^{k + 2}}{3}

(- 3, 2, 2) (1 - t) + (1, - 2, 5) t

(4 + 2 i)^{\circ} (1 + 2 i)

(x + 1) d y

\frac{ln ( x + a ) ( x + a ) + x}{x + a}

(x - 3) \cdot e^{x}

What is expand (7*2^{(k+1)}+(-1)^{(k+2)})/3 ?
The solution to expand (7*2^{(k+1)}+(-1)^{(k+2)})/3 is (7*2^{k+1})/3+((-1)^{k+2})/3