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Popolare Algebra >

espandere (3a+4z)^7

Soluzione

espandere

(3 a + 4 z)^{7}

Soluzione

2187 a^{7} + 20412 a^{6} z + 81648 a^{5} z^{2} + 181440 a^{4} z^{3} + 241920 a^{3} z^{4} + 193536 a^{2} z^{5} + 86016 a z^{6} + 16384 z^{7}

Fasi della soluzione

(3 a + 4 z)^{7}

Applicare il teorema binomiale:

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

a = 3 a, b = 4 z

= i = 0 \sum 7 (i 7) (3 a)^{(7 - i)} (4 z)^{i}

Espandere la sommatoria

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

Semplificare

\frac{7 !}{0 ! ( 7 - 0 ) !} (3 a)^{7} (4 z)^{0} : 2187 a^{7}

Semplificare

\frac{7 !}{1 ! ( 7 - 1 ) !} (3 a)^{6} (4 z)^{1} : 20412 a^{6} z

Semplificare

\frac{7 !}{2 ! ( 7 - 2 ) !} (3 a)^{5} (4 z)^{2} : 81648 a^{5} z^{2}

Semplificare

\frac{7 !}{3 ! ( 7 - 3 ) !} (3 a)^{4} (4 z)^{3} : 181440 a^{4} z^{3}

Semplificare

\frac{7 !}{4 ! ( 7 - 4 ) !} (3 a)^{3} (4 z)^{4} : 241920 a^{3} z^{4}

Semplificare

\frac{7 !}{5 ! ( 7 - 5 ) !} (3 a)^{2} (4 z)^{5} : 193536 a^{2} z^{5}

Semplificare

\frac{7 !}{6 ! ( 7 - 6 ) !} (3 a)^{1} (4 z)^{6} : 86016 a z^{6}

Semplificare

\frac{7 !}{7 ! ( 7 - 7 ) !} (3 a)^{0} (4 z)^{7} : 16384 z^{7}

= 2187 a^{7} + 20412 a^{6} z + 81648 a^{5} z^{2} + 181440 a^{4} z^{3} + 241920 a^{3} z^{4} + 193536 a^{2} z^{5} + 86016 a z^{6} + 16384 z^{7}

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