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

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

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

expand

2 - \frac{k + 2}{2 ^{k}} + \frac{k + 1}{2 ^{k + 1}}

2 - \frac{k}{2 ^{k}} - 2^{1 - k} + \frac{k}{2 ^{k + 1}} + \frac{1}{2 ^{k + 1}}

Solution steps

2 - \frac{k + 2}{2 ^{k}} + \frac{k + 1}{2 ^{k + 1}}

Apply the fraction rule:

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

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

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

Apply the fraction rule:

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

\frac{k + 1}{2 ^{k + 1}} = \frac{k}{2 ^{k + 1}} + \frac{1}{2 ^{k + 1}}

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

Remove parentheses:

(a) = a

= 2 - \frac{k}{2 ^{k}} - \frac{2}{2 ^{k}} + \frac{k}{2 ^{k + 1}} + \frac{1}{2 ^{k + 1}}

Cancel

\frac{2}{2 ^{k}} : 2^{- k + 1}

= 2 - \frac{k}{2 ^{k}} - 2^{- k + 1} + \frac{k}{2 ^{k + 1}} + \frac{1}{2 ^{k + 1}}

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