What is expand 1/pi [+2-(-1)^n+(pi^2)/2-e^pi] ?

The solution to expand 1/pi [+2-(-1)^n+(pi^2)/2-e^pi] is 2/pi-(e^pi)/pi+pi/2-1/pi (-1)^n

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expand 1/pi [+2-(-1)^n+(pi^2)/2-e^pi]

expand

\frac{1}{π} [+ 2 - (- 1)^{n} + \frac{π ^{2}}{2} - e^{π}]

\frac{2}{π} - \frac{e ^{π}}{π} + \frac{π}{2} - \frac{1}{π} (- 1)^{n}

Solution steps

\frac{1}{π} (2 - (- 1)^{n} + \frac{π ^{2}}{2} - e^{π})

Distribute parentheses

= \frac{1}{π} \cdot 2 + \frac{1}{π} (- (- 1)^{n}) + \frac{1}{π} \cdot \frac{π ^{2}}{2} + \frac{1}{π} (- e^{π})

Apply minus-plus rules

+ (- a) = - a

= 2 \cdot \frac{1}{π} - \frac{1}{π} (- 1)^{n} + \frac{1}{π} \cdot \frac{π ^{2}}{2} - e^{π} \frac{1}{π}

Simplify

2 \cdot \frac{1}{π} - \frac{1}{π} (- 1)^{n} + \frac{1}{π} \cdot \frac{π ^{2}}{2} - e^{π} \frac{1}{π} : \frac{2 - e ^{π}}{π} + \frac{π}{2} - \frac{1}{π} (- 1)^{n}

= \frac{2 - e ^{π}}{π} + \frac{π}{2} - \frac{1}{π} (- 1)^{n}

Apply the fraction rule:

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

\frac{2 - e ^{π}}{π} = \frac{2}{π} - \frac{e ^{π}}{π}

= \frac{2}{π} - \frac{e ^{π}}{π} + \frac{π}{2} - \frac{1}{π} (- 1)^{n}

P (6, - 2) y Q (- 8, 4)

ln (4 x) \cdot (x^{2} + x)

(1 - ϕ) \frac{d}{d ϕ} (K (ϕ))

(y^{2} + 7 x y + 16 x^{2}) d x + x^{2} d y

14 (1 + e^{4 + \frac{x}{105}})

What is expand 1/pi [+2-(-1)^n+(pi^2)/2-e^pi] ?
The solution to expand 1/pi [+2-(-1)^n+(pi^2)/2-e^pi] is 2/pi-(e^pi)/pi+pi/2-1/pi (-1)^n