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Popular Álgebra >

expandir (-1/10+2pii)^{10}

Solução

expandir

(- \frac{1}{10} + 2 πi)^{10}

Solução

(- 1024 π^{10} + \frac{1}{1 0 ^{10}} + \frac{576 π ^{8}}{5} + \frac{21 π ^{4}}{6250} - \frac{168 π ^{6}}{125} - \frac{9 π ^{2}}{5000000}) + i \frac{- π - 25600000000 π ^{9} - 4032000 π ^{5} + 4800 π ^{3} + 768000000 π ^{7}}{50000000}

Passos da solução

(- \frac{1}{10} + 2 πi)^{10}

Aplicar o teorema do binômio:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = - \frac{1}{10}, b = 2 πi

= i = 0 \sum 10 (i 10) (- \frac{1}{10})^{(10 - i)} (2 πi)^{i}

Expandir a soma

= \frac{10 !}{0 ! ( 10 - 0 ) !} (- \frac{1}{10})^{10} (2 πi)^{0} + \frac{10 !}{1 ! ( 10 - 1 ) !} (- \frac{1}{10})^{9} (2 πi)^{1} + \frac{10 !}{2 ! ( 10 - 2 ) !} (- \frac{1}{10})^{8} (2 πi)^{2} + \frac{10 !}{3 ! ( 10 - 3 ) !} (- \frac{1}{10})^{7} (2 πi)^{3} + \frac{10 !}{4 ! ( 10 - 4 ) !} (- \frac{1}{10})^{6} (2 πi)^{4} + \frac{10 !}{5 ! ( 10 - 5 ) !} (- \frac{1}{10})^{5} (2 πi)^{5} + \frac{10 !}{6 ! ( 10 - 6 ) !} (- \frac{1}{10})^{4} (2 πi)^{6} + \frac{10 !}{7 ! ( 10 - 7 ) !} (- \frac{1}{10})^{3} (2 πi)^{7} + \frac{10 !}{8 ! ( 10 - 8 ) !} (- \frac{1}{10})^{2} (2 πi)^{8} + \frac{10 !}{9 ! ( 10 - 9 ) !} (- \frac{1}{10})^{1} (2 πi)^{9} + \frac{10 !}{10 ! ( 10 - 10 ) !} (- \frac{1}{10})^{0} (2 πi)^{10}

\frac{10 !}{0 ! ( 10 - 0 ) !} (- \frac{1}{10})^{10} (2 πi)^{0} = \frac{1}{1 0 ^{10}}

Simplificar

\frac{10 !}{1 ! ( 10 - 1 ) !} (- \frac{1}{10})^{9} (2 πi)^{1} : - \frac{πi}{50000000}

\frac{10 !}{2 ! ( 10 - 2 ) !} (- \frac{1}{10})^{8} (2 πi)^{2} = - \frac{9 π ^{2}}{5000000}

Simplificar

\frac{10 !}{3 ! ( 10 - 3 ) !} (- \frac{1}{10})^{7} (2 πi)^{3} : \frac{3 π ^{3} i}{31250}

\frac{10 !}{4 ! ( 10 - 4 ) !} (- \frac{1}{10})^{6} (2 πi)^{4} = \frac{21 π ^{4}}{6250}

Simplificar

\frac{10 !}{5 ! ( 10 - 5 ) !} (- \frac{1}{10})^{5} (2 πi)^{5} : - \frac{252 π ^{5} i}{3125}

\frac{10 !}{6 ! ( 10 - 6 ) !} (- \frac{1}{10})^{4} (2 πi)^{6} = - \frac{168 π ^{6}}{125}

Simplificar

\frac{10 !}{7 ! ( 10 - 7 ) !} (- \frac{1}{10})^{3} (2 πi)^{7} : \frac{384 π ^{7} i}{25}

\frac{10 !}{8 ! ( 10 - 8 ) !} (- \frac{1}{10})^{2} (2 πi)^{8} = \frac{576 π ^{8}}{5}

Simplificar

\frac{10 !}{9 ! ( 10 - 9 ) !} (- \frac{1}{10})^{1} (2 πi)^{9} : - 512 π^{9} i

\frac{10 !}{10 ! ( 10 - 10 ) !} (- \frac{1}{10})^{0} (2 πi)^{10} = - 1024 π^{10}

= \frac{1}{1 0 ^{10}} - \frac{πi}{50000000} - \frac{9 π ^{2}}{5000000} + \frac{3 π ^{3} i}{31250} + \frac{21 π ^{4}}{6250} - \frac{252 π ^{5} i}{3125} - \frac{168 π ^{6}}{125} + \frac{384 π ^{7} i}{25} + \frac{576 π ^{8}}{5} - 512 π^{9} i - 1024 π^{10}

Agrupar termos semelhantes

= - 1024 π^{10} - 512 π^{9} i + \frac{1}{1 0 ^{10}} + \frac{576 π ^{8}}{5} - \frac{πi}{50000000} + \frac{21 π ^{4}}{6250} - \frac{168 π ^{6}}{125} + \frac{384 π ^{7} i}{25} + \frac{3 π ^{3} i}{31250} - \frac{9 π ^{2}}{5000000} - \frac{252 π ^{5} i}{3125}

Reescrever

- 1024 π^{10} - 512 π^{9} i + \frac{1}{1 0 ^{10}} + \frac{576 π ^{8}}{5} - \frac{πi}{50000000} + \frac{21 π ^{4}}{6250} - \frac{168 π ^{6}}{125} + \frac{384 π ^{7} i}{25} + \frac{3 π ^{3} i}{31250} - \frac{9 π ^{2}}{5000000} - \frac{252 π ^{5} i}{3125}

na forma complexa padrão:

(- 1024 π^{10} + \frac{1}{1 0 ^{10}} + \frac{576 π ^{8}}{5} + \frac{21 π ^{4}}{6250} - \frac{168 π ^{6}}{125} - \frac{9 π ^{2}}{5000000}) + \frac{- 25600000000 π ^{9} - π + 768000000 π ^{7} + 4800 π ^{3} - 4032000 π ^{5}}{50000000} i

= (- 1024 π^{10} + \frac{1}{1 0 ^{10}} + \frac{576 π ^{8}}{5} + \frac{21 π ^{4}}{6250} - \frac{168 π ^{6}}{125} - \frac{9 π ^{2}}{5000000}) + \frac{- 25600000000 π ^{9} - π + 768000000 π ^{7} + 4800 π ^{3} - 4032000 π ^{5}}{50000000} i

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