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Popular Álgebra >

expandir (2e+f)^7

Solução

expandir

(2 e + f)^{7}

Solução

f^{7} + 14 e f^{6} + 84 e^{2} f^{5} + 280 e^{3} f^{4} + 560 e^{4} f^{3} + 672 e^{5} f^{2} + 448 e^{6} f + 128 e^{7}

Passos da solução

(2 e + f)^{7}

Aplicar o teorema do binômio:

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

a = 2 e, b = f

= i = 0 \sum 7 (i 7) (2 e)^{(7 - i)} f^{i}

Expandir a soma

= \frac{7 !}{0 ! ( 7 - 0 ) !} (2 e)^{7} f^{0} + \frac{7 !}{1 ! ( 7 - 1 ) !} (2 e)^{6} f^{1} + \frac{7 !}{2 ! ( 7 - 2 ) !} (2 e)^{5} f^{2} + \frac{7 !}{3 ! ( 7 - 3 ) !} (2 e)^{4} f^{3} + \frac{7 !}{4 ! ( 7 - 4 ) !} (2 e)^{3} f^{4} + \frac{7 !}{5 ! ( 7 - 5 ) !} (2 e)^{2} f^{5} + \frac{7 !}{6 ! ( 7 - 6 ) !} (2 e)^{1} f^{6} + \frac{7 !}{7 ! ( 7 - 7 ) !} (2 e)^{0} f^{7}

\frac{7 !}{0 ! ( 7 - 0 ) !} (2 e)^{7} f^{0} = 128 e^{7}

Simplificar

\frac{7 !}{1 ! ( 7 - 1 ) !} (2 e)^{6} f^{1} : 448 e^{6} f

Simplificar

\frac{7 !}{2 ! ( 7 - 2 ) !} (2 e)^{5} f^{2} : 672 e^{5} f^{2}

Simplificar

\frac{7 !}{3 ! ( 7 - 3 ) !} (2 e)^{4} f^{3} : 560 e^{4} f^{3}

Simplificar

\frac{7 !}{4 ! ( 7 - 4 ) !} (2 e)^{3} f^{4} : 280 e^{3} f^{4}

Simplificar

\frac{7 !}{5 ! ( 7 - 5 ) !} (2 e)^{2} f^{5} : 84 e^{2} f^{5}

Simplificar

\frac{7 !}{6 ! ( 7 - 6 ) !} (2 e)^{1} f^{6} : 14 e f^{6}

Simplificar

\frac{7 !}{7 ! ( 7 - 7 ) !} (2 e)^{0} f^{7} : f^{7}

= 128 e^{7} + 448 e^{6} f + 672 e^{5} f^{2} + 560 e^{4} f^{3} + 280 e^{3} f^{4} + 84 e^{2} f^{5} + 14 e f^{6} + f^{7}

Reescrever na forma geral

= f^{7} + 14 e f^{6} + 84 e^{2} f^{5} + 280 e^{3} f^{4} + 560 e^{4} f^{3} + 672 e^{5} f^{2} + 448 e^{6} f + 128 e^{7}

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