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Popular Álgebra >

simplificar (2+i)^7

Solução

simplificar

(2 + i)^{7}

Solução

- 278 - 29 i

Passos da solução

(2 + i)^{7}

Aplicar o teorema do binômio:

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

a = 2, b = i

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

Expandir a soma

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

Simplificar

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 2^{7} i^{0} : 128

Simplificar

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 2^{6} i^{1} : 448 i

Simplificar

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 2^{5} i^{2} : 672 i^{2}

Simplificar

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 2^{4} i^{3} : 560 i^{3}

Simplificar

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 2^{3} i^{4} : 280 i^{4}

Simplificar

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 2^{2} i^{5} : 84 i^{5}

Simplificar

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 2^{1} i^{6} : 14 i^{6}

Simplificar

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 2^{0} i^{7} : i^{7}

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

Simplificar

128 + 448 i + 672 i^{2} + 560 i^{3} + 280 i^{4} + 84 i^{5} + 14 i^{6} + i^{7} : - 29 i - 278

= - 29 i - 278

Rewrite in standard complex form:

- 278 - 29 i

= - 278 - 29 i

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