expandir-(2l((-1)^n-1))/(pi^2n^2)

Solução

expandir

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

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

Passos da solução

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

Expandir

2 l ((- 1)^{n} - 1) : 2 (- 1)^{n} l - 2 l

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

Aplicar as propriedades das frações:

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

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

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

Remover os parênteses:

(a) = a, - (- a) = a

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

e x p an d \frac{2 ( 1 + x ) y}{( 1 - 2 x - x ^{2} )}

e x p an d \frac{3 ( 3 y - 1 )}{- 3 x + 2}

e x p an d (x^{2} - 4 x) d x 1. 3^{^{'}}

e x p an d \frac{1}{x ( x + 2 ) ( x - 3 )}

e x p an d (e^{2 k x} + e^{- k x})^{5}