expandir (2a+3b)^4

Solução

expandir

(2 a + 3 b)^{4}

16 a^{4} + 96 a^{3} b + 216 a^{2} b^{2} + 216 a b^{3} + 81 b^{4}

Passos da solução

(2 a + 3 b)^{4}

Aplicar o teorema do binômio:

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

a = 2 a, b = 3 b

= i = 0 \sum 4 (i 4) (2 a)^{(4 - i)} (3 b)^{i}

Expandir a soma

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2 a)^{4} (3 b)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2 a)^{3} (3 b)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2 a)^{2} (3 b)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2 a)^{1} (3 b)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2 a)^{0} (3 b)^{4}

Simplificar

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 a)^{4} (3 b)^{0} : 16 a^{4}

Simplificar

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 a)^{3} (3 b)^{1} : 96 a^{3} b

Simplificar

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 a)^{2} (3 b)^{2} : 216 a^{2} b^{2}

Simplificar

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 a)^{1} (3 b)^{3} : 216 a b^{3}

Simplificar

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 a)^{0} (3 b)^{4} : 81 b^{4}

= 16 a^{4} + 96 a^{3} b + 216 a^{2} b^{2} + 216 a b^{3} + 81 b^{4}

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